Class – 11 Physics, Chapter – 6, Motion in a Plane

Projectile Motion : 

Questions on this topics :

1 ) Define projectile or what do you mean by projectile?

2 ) Give examples of projectile.

3 ) Define trajectory.
             or 
    what is called the path followed by a projectile?

4 ) what are the assumption regarding the projectile motion?

5 ) There are how many projectiles? Name them.

6 ) what is the principle of physical independence of motion?

7 ) what is the nature of trajectory of horizontal projectile?

8 ) Is the horizontal motion of a projectile uniform motion? If yes, why?

9 ) The vertical motion of a projectile is controlled by which type of force?

10 ) what is the trajectory of horizontal projectile?

11 ) derive the expression for time of flight of a horizontal projectile?

12 ) what do you mean by horizontal range of a horizontal projectile? Derive its
     expression.

13 ) Derive the expression of trajectory of oblique projectile. 
                          or 
     Prove that the trajectory of an oblique projectile is parabola.

14 ) Derive the expression for the resultant velocity of oblique projectile.

15 ) Derive the expression for maximum height of oblique projectile.

16 ) Derive the expression for horizontal range of oblique projectile.

17 ) There are how many angles of projection for the same range? Derive its 
     expression.

Now let us discuss answers of above questions.

[ Dear students, 
      You will get all answers of above questions but you have to search. 
The quetions are written first so that you may prepare yourself to study about 
this topic.]

Definition of projectile :

A body which is in flight through the atmosphere but is not being propelled by any fuel is called a projectile. The motion of the projectile is termed as projectile motion.

Examples : i) a bullet fired from a gun        ii) an arrow released from a bow       iii) golf ball in flight        iv) a bus driven off an uncompleted bridge.

Trajectory : The path followed by a projectile is called as trajectory. The nature of trajectory of a projectile is parabolic due to following reason

i ) The vertical position of the projectile is influenced only by a constant acceleration.

ii ) Horizontal velocity of a projectile is generally constant.

Condition of projectile motion :

Projectile motion only occurs when there is one force applied at the beginning, after which the only influence on the trajectory is that of gravity.

Assumptions regarding projectile motion : 

i) There is no air resistance.

ii) The effect due to earth curvature is negligible.

iii) The effect of rotation of Earth is negligible.

iv) For all points of trajectory, the acceleration due to gravity is constant in mag- tude and direction.

Types of projectile :

There are two types of projectile.

i) Horizontal projectile

ii) Oblique projectile

Horizontal projectile :

a ) Definition : If a body is projected horizontally from a certain height with a certain velocity , then the body is called as horizontal projectile.

b ) Examples :

i) a ball is thrown from roof. ii) A jet of water from a hole near the bottom of a water tank. iii) A glass accidentally falling off a table.

c ) Trajectory : 

The trajectory of a projectile is parabola.

Proof :

Let a projectile is thrown from a certain height with certain velocity u. The projectile motion is a two dimensional motion. let us discuss it in two part

For horizontal motion,

initial velocity = u

horizontal displacement = x

As there is only force of gravity acting on the projectile, so

horizontal component of acceleration = 0

equation of motion is given by

x = ut + 1/2 at²

⇒ x = ut + 0

⇒ t = x / u  ———————————- ( 1 )

For vertical motion,

initial velocity = 0

vertical displacement = y

vertical component of acceleration = + g

where g is acceleration due to gravity

equation of motion is given by

y = 0*t + 1/2 gt²

⇒ y = 1/2 * g * ( x / u ) ²

⇒ y = [ g / ( 2u² ) ] * x²

⇒ y = kx²   ———————————- ( 2 )

where k = [ g / ( 2u² ) ]

equation ( 2 ) represents the equation of parabola. hence, the trajectory of a projectile is parabola.

d ) Time of flight  ( T )_:

It is the time of descending of a projectile from point of projection to the ground.

Let h = vertical height of point of projection above the ground

g = acceleration due to gravity

T = time of flight

initial velocity ( u ) = 0

vertical component of acceleration ( a ) = + g

vertical displacement ( s ) = h

Let us consider the vertical motion, the equation of motion is given by

s = ut + 1/2at²

⇒       h = 0 + 1/2gT²

T = √( 2h / g)

e ) Horizontal range  ( R  ) :

It is the horizontal distance traveled by projectile during the time of flight.

Let time of flight ( t ) = T = √( h / 2g )

horizontal displacement ( s ) = R

horizontal component of acceleration ( a ) = 0

initial velocity = u

For horizontal motion, the equation of motion is given by

s = ut + 1/2at²

R = u √( 2h / g)

f ) Instantaneous velocity (v) :

Instantaneous velocity of a projectile is the resultant velocity at any point on its trajectory at any instant of time.

Let v = resultant velocity at any point A on the trajectory

t = time to reach at point A from point of projection

v_y = vertical component of v

since, the horizontal motion of the projectile is uniform motion. Therefore,

v_x = horizontal component of v = u = initial velocity

+g = vertical component of acceleration

but horizontal component of acceleration is zero as horizontal motion is uniform.

for vertical motion, the equation of motion is given by

v_y = u + at

= 0 + gt

= gt ——————————————- ( 1 )

resultant velocity ( v ) = √ ( v²_x + v²_y )

v = √ ( u² + g²t² )

let θ = angle made by resultant velocity with its horizontal component 

therefore,   tanθ = v_x / v_y

= u / gt

θ = tan^-1 ( u / gt )

Oblique projectile :

a) Definition : If a body is projected at a certain angle with the horizontal, then the body is called as oblique projectile.

b) Trajectory :

Let a projectile is thrown with a certain velocity u at a certain angle Ф  with the horizontal.

For horizontal motion ,

Initial velocity ( u_x ) = ucosФ

Horizontal motion is uniform motion, so

horizontal component of acceleration ( a_x ) = 0

Horizontal displacement ( s_x ) = x

The equation of motion is given by

s_x = u_xt + 1/2 a_xt²

⇒ x = ucosФ t + 0

⇒ t = x / ucosФ  ————————— ( 1 ) 

For vertical motion,

initial velocity ( u_y ) = usinФ

vertical component of acceleration ( a_y ) = -g

vertical displacement ( s_y ) = y

The equation of motion is given by

s_y = u_yt + 1/2 a_yt²

⇒ y = usinФ t + ( – g ) t² / 2

⇒ y = usinФ( x / ucosФ ) – g ( x / ucosФ )² / 2                  [ using equation ( 1 ) ]

⇒ y = x tanФ – [g / ( 2u²cos²Ф ) ] * x²

Above equation represents the equation of parabola

Hence, the trajectory of oblique projectile is parabolic in nature.

c) Time of flight :

let time of flight = T

time taken to reach at the highest point = t

∴ T = 2t 

For the motion at the highest point ,

vertical component of  initial velocity ( u ) = usinФ

vertical component of acceleration ( a ) = – g

final velocity ( v ) = 0

The equation of motion is given by

v = u + at

⇒ 0 = usinФ + ( – gt ) 

⇒ usinФ = gt

⇒ t = usinФ / g

∴ T = 2t 

T = 2usinФ / g

d) Horizontal range :

For horizontal motion,

initial velocity ( u_x ) = ucosФ 

time of flight ( T ) = 2usinФ / g

horizontal displacement = horizontal range ( s ) = R

horizontal component of acceleration ( a ) = 0

The equation of motion is given by

s = u_xT + 1/2 aT²

R = ucosФ ( 2usinФ / g ) + 0

= ( 2cosФsinФ )u² / g

R = u² sin2Ф / g

for maximum horizontal range

sin2Ф = 1

⇒ sin2Ф = sin 90

⇒ 2Ф = 90

⇒ Ф = 45 

e) Maximum Height ( H ) :

at the highest point

initial velocity ( u_y ) = usinФ

final velocity  ( v_y ) = 0

vertical component of acceleration ( a_y ) = – g

the equation of motion is given by

v_y = u_y + a_yt

⇒ 0 = usinФ + ( – g ) t 

⇒ t = usinФ / g —————————– ( 1 ) 

and    s_y = u_yt + 1 / 2a_yt²

H = usinФ ( usinФ / g ) +  1/2 ( – g ) ( usinФ / g )²

= ( u²sin²Ф / g ) – u²sin²Ф / 2g

= u²sin²Ф / 2g

for maximum height

sinФ = 1

⇒  sinФ = sin90

⇒  Ф = 90

H_max = u² / 2g

 This maximum height is also termed as vertical range.

Principal of physical independence of motion : 

The projectile motion is a two dimensional motion. So, it can be discussed in two parts. One is horizontal motion and other is vertical motion. These two motions takes place independently to each other. This is termed as principal of physical independence of motion.
Important points :

i) A projectile motion is 2-dimensional motion.

ii) The trajectory of the projectile motion is parabola provided air resistance, effect of earth curvature and rotation of the earth are negligible.

iii) horizontal motion of a projectile is uniform because the only force acting on it is force of gravity ( acting vertically downward and its horizontal compo -nents is zero. )

iv) The vertical motion of a projectile is controlled by force of gravity and is accelerated motion.

v) The initial velocity in vertically downward direction is zero.

vi) Mass does not affect the speed of falling objects, assuming there is only gravity acting on it. Both bullets will strike the ground at the same time. The horizontal force applied does not affect the downward motion of the bullets – only gravity and friction (air resistance), which is the same for both bullets.

Factors affecting the flight path of a Projectile are:

i) Gravity.

ii) Air Resistance.

iii) Speed of Release.

iv) Angle of Release.

v) Height of Release.

vi) Spin.

Two angles of projection for the same range :

R = u² sin2Ф / g

= u²sin²( 180 – 2Ф ) / g

R = u² {sin2( 90 – Ф ) }²/ g 

Hence, there are two angles of projection Ф and ( 90 – Ф ) for the same horizontal range R.

This means that the horizontal range R is same whether the projectile is thrown at an angle Ф with the horizontal or at an angle Ф with the vertical.

Numerical problems 

1 ) A dart is thrown horizontally towards the bull’s eye point P on the dart board of the following figure with an initial speed  of 10 m/sec. It hits at point Q on the rim, vertically below P, 0.19 sec later. (a) what is the distance PQ ? (b) How far away from the dart board did the dart thrower stand?

Solution : 

( a) for vertical motion, initial velocity is zero

equation of motion is given by

s_y = u_yt + 1/2a_yt²

= 0 + 1/2 *9.8*0.19*0.19

= 0.176 meter

required distance = PQ = 0.176 meter

( b ) for horizontal motion, initial velocity ( u ) = 10 m/sec

and acceleration is zero

equation of motion is given by

s = ut + 1/2at²

=  10*0.19 + ( 1/2*0*0.19*0.19 )

= 1.9 meter

2 ) A projectile is fired with a horizontal velocity of 330 m/sec from the top of a cliff 80 m high. How long will it take to strike the level ground at the base of cliff ? With what velocity it strike ? Neglect air resistance.

Solution : 

Initial velocity ( u ) = 330 m/sec

Height ( h ) = 80 m

acceleration ( a ) = acceleration due to gravity = 9.8 m/sec²

required time = time of flight = T = √( 2h/g)

= √{ (2*80)/9.8 }

= 4.04 sec

For vertical motion,

v_y = u_y + a_yt

= 0 + ( -9.8* 4.04)

= – 39.59

required velocity= – 39.59 cm/sec

( negative sign indicates that the velocity is in downward direction )

3 ) A stunt flier is flying at 15 m/sec parallel to the flat ground 100 m below. How large must be the distance x of the plane is to hit the target?

Solution :

initial velocity ( u ) = 15 m/sec

height ( h ) = 100 m

acceleration due to gravity( g ) =9.8 m/sec²

x = u√(2h/g)

= 15*√(2*100/9.8)

= 67.76 m

4 ) A projectile is thrown at an angle θ with the horizontal with kinetic energy E. Calculate the potential energy at the top most point of trajectory.

Solution : 

kinetic energy = E

⇒ 1/2 mu² = E  —————————————- ( 1 )

potential energy = mgh

where h is the maximum height

h = u²sin²θ / 2g

Therefore, potential energy = mgh

=  mg( 1/2 m²u² ) sinθ / g

=  mEsin²θ

 5 ) A projectile is thrown with an initial velocity of xî + yĵ. The range of projectile is twice the maximum height of projectile. Calculate y / x.

Solution : 

initial velocity ( u ) = xî + yĵ

where x = ucosθ

y = usinθ

Range ( R ) = 2* maximum height

⇒ u²sin2θ / g = 2 * u²sin²θ / 2g

⇒ u² ( 2sinθcosθ ) / g = 2 * u²sin²θ / 2g

⇒ 2 ( ucosθ ) ( usinθ ) = ( usinθ ) ( usinθ ) 

⇒ 2xy = y*y

⇒ 2x  = y

⇒ y / x =2

6 ) A projectile has a range of 50 m and reaches a maximum height of 10 m. What is the elevation of the projectile ?

Solution :

angle of elevation ( θ ) = ???

Range ( R ) = 50 m

⇒ u²sin2θ / g  = 50

⇒ u² = 50g / sin2θ   ———————————- ( 1 )

Maximum height = 10 m

u²sin²θ / 2g = 10

{50g / sin2θ}sin²θ/ 2g = 10

50sin²θ / ( 4gsinθcosθ ) = 10

tanθ = 4g / 5

θ = tan^-1 { ( 4*9.8) / 5 }

θ = tan^-1 ( 7.84 )

θ = 82.73 degree

7 ) A cricket ball is thrown at a speed of 28 m/sec in a direction 30 degree above the horizontal. Calculate (a) the maximum height, (b) the time taken by ball to return to same level, and (c) the distance from the thrower to the point where the ball returns to same level.

Solution :

Initial velocity  ( u ) = 28 m/sec

angle of elevation ( θ ) = 30 degree

acceleration due to gravity ( g ) = 9.8 m/sec²

( a ) Maximum height = u²sin²θ / 2g

= 28² * sin²30  / ( 2 * 9.8 )

= 10 m

( b ) The time taken by ball to return to same level = time of flight

= 2usinθ / g

= ( 2 * 28 * sin30 ) / g

= 2.86 sec

(c) the distance from the thrower to the point where the ball returns to same level = horizontal range = u²sin2θ / g

= { 28*28*sin( 2*30  ) } / 9.8

= 69.2 m

 8 ) From the same point, two balls A and B are thrown simultaneously. A is thrown vertically up with a velocity of 20 m/sec. B is thrown with a velocity of 20 m/sec at an angle of 60 degree with the vertical. Determine the separation between the two balls at t = 1 sec.

Solution : Let  x = horizontal separation

y = vertical separation

for horizontal motion, the equation of motion is given by

x  = ucos( 90 – 60 )t + 1/2at²

= 20* cos30 *1 + 0

= 10√3

for vertical motion , the equation of motion is given by

y = ( usin90t + 1/2at² ) – ( usin( 90 – 60 )* 1 – 1/2 at² ) 

y = ( 20 *1 – 9.8/2 ) – ( 20sin30 – 9.8 / 2 )

y = 20 – 4.9 – 10 + 4.9

y = 10 m

required separation = √ ( x² + y² )

= √ { ( 10√3 )² +( 10 )² }

= √400

= 20 m

9 ) A projectile is thrown with velocity v at angle θ with the horizontal. Its maximum height is 8 m and horizontal is 24 m. Calculate v in terms of acceleration due to gravity g. Also determine the sinθ . 

Solution :

initial velocity ( u ) = v

maximum height ( H ) = 8 m

horizontal range ( R ) = 24 m

we know that

H = u²sin²θ / 2g ————————— ( 1 ) 

R = u²sin2θ / g  ——————————— ( 2 ) 

( 1 ) ÷ ( 2 )

H / R = sin²θ / 4sinθcosθ

8 / 24 = tanθ / 4

tanθ = 3/4

θ = 36.87 degree

Now putting the value of θ in equation ( 1 ) we get 

H = u²sin²θ / 2g

8 = v sin(36.87) /2g

v = ( 8 * 2g ) / sin(36.87)

= 26.67g

10 ) A ball is thrown with an initial velocity of 100 m/sec at an angle of 30 degree above the horizontal. How far from the throwing point will the ball attain its original level ? Solve the problem without using formula for horizontal range ?

Solution :

initial velocity ( u ) = 100 m/sec

angle of projection ( θ ) = 30 degree

time of flight ( T ) = 2usinθ / g = ( 2*100*sin30 ) / 9.8 = 10.204 sec

for horizontal motion, the equation of motion is given by

R = ucos30T + 1/2 aT²

= 100*cos30*10.204 + ( 0 * 10.204*10.204 ) / 2

= 883.7 m

11 ) Prove that the velocity at the end of flight of an oblique projectile is the same in magnitude at the beginning but the angle that its with the horizontal is negative of the angle of projection.

Solution : 

Let initial velocity = u 

velocity at the end = v 

angle of projection = θ

time of flight ( T ) = 2usinθ / g

For horizontal motion at the end of flight, the equation of motion is given by 

v_x = u_x + a_xt

as the horizontal motion is uniform motion, the horizontal component of acceleration is zero.

v_x = ucosθ + 0

     = ucosθ  ————————————- ( 1 )

For vertical motion at the end of flight, the equation of motion is given by 

v_y = usinθ + a_yT

     = usinθ – g*( 2usinθ / g )

     = – usinθ

Therefore, 

v = √ ( v²_x + v²_y )

   = √ {( ucosθ )² + ( – usinθ )² }

   = √ { u² ( cos²θ + sin²θ )

   = u

Hence, the velocity at the end of flight of an oblique projectile is the same in magnitude at the beginning but the angle that its with the horizontal is negative of the angle of projection.

( Proved ) 

12 ) A particle starts from origin at equal t = 0 with a velocity 5.0 î m/sec and moves x-y plane under action of force which produce a constant acceleration of ( 3.0 î + 2.0 ĵ ) m/sec². (a) what is the y- coordinate of the particle at the instant its x – coordinate 84 m ? (b) what is the speed of the particle at this time ?

Solution : 

initial velocity ( u ) = 5.0 î m/sec

acceleration ( a ) = ( 3.0 î + 2.0 ĵ ) m/sec²

 x – coordinate = 84 m  m/sec²

y – coordinate = ??

The equation of motion is given by 

r = u t + at² / 2

   =  5.0 î t + ( 3.0 î + 2.0 ĵ )t²

   = (5.0t + 1.5t² )î + 1.0t²ĵ

   x – coordinate = 84 m 

   (5.0t + 1.5t² ) = 84

    on solving we get t = 6 sec 

y-coordinate = 1.0t² = 1.0*6*6 = 36 m 

Required velocity = v

  v = dr / dt

      = ( 5.0 + 3.0t )î + 2.0tĵ

      putting t = 6 sec 

v = 23î + 2 ĵ

v = √ ( 23² + 2² ) = 23.09 m/sec 

13 ) The position of a particle is given by r = 3.0t i + 2.0 t² j + 5.0 k  where t is in second and the coefficients have the proper units for r to be in meter. (a) Find v(t) and a(t) of the particle. (b) Find the magnitude and direction of v(t) at t = 3.0 sec.

Solution :

( a )

r = 3.0t î + 2.0 t² ĵ+ 5.0 k̂ 

v = dr / dt

= 3.0î + 4.0tĵ + 0k̂ = 3.0î + 4.0tĵ

a = dv / dt = 0 + 4.0ĵ = 4.0ĵ

( b )

v(t = 3.0 ) = 3.0î + 12.0ĵ

magnitude of v  at t = 3.0 sec is given by

v = √ ( 3² + 12² ) = 12.37 m/sec

Let θ is the angle made by y – component  of v with x – axis 

tanθ = ( y – component of v ) / ( x – component of v ) = 12.0 / 3.0 = 4.0

θ = tan^-1 ( 4.0 ) = 75.97 degree

14 ) An aeroplane moving horizontally at 20 m/sec drops a bag. What is the displacement of the bag after 5 sec ?                     [ RPMT 1993 ]

Solution :

initial velocity ( u ) = 20 m / sec

time ( t ) = 5 sec

horizontal displacement after 5 sec = x

vertical  displacement after 5 sec = y

required displacement = r = √( x² + y² )

for horizontal motion is given by

x = ut + 1/2 at²

= ( 20*5 ) + 0 = 100 m

The equation of trajectory is given by

y = [ g / ( 2u² ) ] * x²

= [ 9.8 / ( 2*20*20 ) ] *100*100 = 122.5 m

r = √( x² + y² ) = √ ( 100² + 122.5² ) = 158.13 m

15 ) A ball is projected horizontally with a velocity of 3 m/sec . What is velocity of the ball after 0.4 sec ? given : g = 10 m/sec                               [ Bihar PMT 1997, modified ]

Solution :

initial velocity ( u ) = 3 m/sec

velocity after 0.4 sec = √ ( u² + g²t² ) = √ { 3² +( 10*10*0.4*0.4 ) } = 5 m / sec

16) From the top of a wall of height 20 m , a ball is thrown horizontally with a speed of 6 m/sec . How far from the wall will the ball land ?

Solution :

initial velocity ( u ) = 6 m /sec

height ( h ) = 20

distance from the wall will the ball land = horizontal range = u √ ( 2h / g ) = 6√ ( 40 / 9.8 ) = 12.122 m  

17 ) A projectile is fired with an initial speed of 500 m/sec horizontally from the top of a cliff of height 19.6 m. At what distance from the foot of the cliff, does it strike the ground ?

Solution :

initial speed ( u ) = 500 m /sec

height ( h ) = 19.6 m

required distance = horizontal range = u √ ( 2h / g ) =500√ { ( 2*19.6 ) / 9.8 } = 1000 m 

18 ) A marble with with speed 20 cm /sec rolls off the edge of a table 80 cm high.  (a) How long does it take to drop to the floor ? (b) How far horizontally from the table edge does the marble strike the marble ?

Solution :

initial speed ( u ) = 20 cm /sec

height ( h ) = 80 cm

( a )  required time = time of flight = √ ( 2h / g ) = √ ( 160 / 20 ) = 2.83 sec 

( b ) required distance = horizontal range = u √ ( 2h / g ) = 20 √ ( 160 / 20 ) = 56.6 cm  

19 ) A body of mass m is thrown horizontally with a velocity of 60 km/hr from the top of the tower of height h. It touches the level ground at a distance of 400 m from the foot of the tower. Now, a body of mass 2m is thrown horizontally with a velocity of 30 km/hr from the top of a tower of height 4h. At what distance from the foot of the tower would it touch the level ground ?

Solution :

For body of mass m : 

initial speed ( u ) = 60 km / hr = 16.67 m/sec

height = h

horizoninitial velovcity al range ( R ) = 400 m

u√( 2h / g ) = 400

squaring both sides we get

2hu² / g = 160000

h = 160000g / 2u²

h = ( 160000 * 9.8 ) / {2 * ( 16.67 )² }

h = 2821.3 m = 2.82 km

For body of mass 2m:

initial velocity ( u ) = 30 km / hr

height ( H ) = 4h = 4*2.82 = 11.28 km / hr

required distance = horizontal range = u√( 2H / g ) = 30 * √ ( 2 * 11.28 / 9.8 ) = 46.55 km

20 ) A shell is fired horizontally from a gun situated at a height of 44.1 m above a horizontal plane with a muzzle velocity of 300 m/sec to to hit a target on the horizontal plane. (a) How long does the shell take to hit the target ? (b) where is the target situated ? (c) With what vertical velocity does the shell strike the target ? Given : g = 9.8 m/sec²

Solution :

initial velocity ( u ) = 300 m / sec

height ( h ) = 44.1 m

( a ) required time = time of flight ( T ) = √( 2h / g ) = √ ( 2*44.1 / 9.8 ) = 3 sec

( b ) horizontal range ( R ) = uT = 300 * 3 = 900 m / sec

( c ) required velocity = acceleration due to gravity * time of flight = 9.8 * 3 = 29.4 m / sec

21 ) An aeroplane is flying in a horizontal direction with a velocity of 360 km/hr at a height of 1960 m . How far from a given target should it release a bomb so as to hit the target ? Given : g = 9.8 m/sec²

Solution :

initial velocity ( u ) = 360 km/hr = 100 / sec

height ( h ) = 1960 m

acceleration due to gravity ( g ) = 9.8 m/sec²

horizontal range ( R ) = u√( 2h / g ) = 100 * √( 2*1960 / 9.8 ) = 20* 100 = 2000 m

22 ) A hiker stands on the edge of of a cliff 490 m above the ground and throws stone horizontally with an initial speed of 15 m/sec . Neglecting air resistance, find the time taken by the stone to reach the ground, and the speed with which it hits the ground. Take g = 9.8 m / sec²

Solution : 

height ( h ) = 490 m

initial speed ( u ) = 15 m / sec

required time ( T ) = √( 2h / g ) = √{(2*490 ) / 9.8 } = 10 sec

required velocity = acceleration due to gravity * time of flight = 9.8 * 10 = 9.8 m /sec

23 ) A projectile is thrown horizontally from the top of tower and strikes the ground after 3 second at an angle of 45 degree with the horizontal. Find the height of the tower and the speed with the horizontal. Find the height of the tower and the speed with which the body was projected. Given : g = 9.8 m/sec² 

Solution :

time of flight ( T ) = 3 sec

let

height = h

according to problem

T = √( 2h / g )

squiring both sides  we get

T² = 2h / g

3² = 2h / 9.8

9 * 9.8 = 2h

88.2 / 2 = h

h= 44.1 m

required velocity = acceleration due to gravity * time of flight = 9.8 * 3 = 29.4 m / sec

24 ) A projectile thrown from a horizontal plane reaches back the plane after covering a horizontal distance of 40 m. If the horizontal velocity of the projectile is 10 m/sec , then what is its initial vertical velocity ? Given : g = 10 m / sec²

Solution : 

horizontal range ( R ) = 40 m

let initial velocity = u

initial horizontal velocity = 10 m /sec

ucosθ = 10 —————————– ( 1 ) 

and R = 40

2( usinθ )( ucosθ ) / g = 40

2 * ( usinθ ) * 10 = 40 * 10 

usinθ  = 400 / 20 

usinθ = 20

initial vertical velocity = 20 m /sec

25 ) A hose lying on the ground shoots a stream of water upwards at an angle of 40 degree to the horizontal. The speed of water is 20 m/sec as it leaves the hose. How high up will it strike a wall which is 8 meter away ?

Solution : 

angle of projection ( θ ) = 40 degree

initial speed ( u ) = 20 m / sec

horizontal displacement ( x ) = 8 m 

required distance = vertical displacement ( y ) = ???  

Equation of trajectory is given by

y = xtanθ  – gx² / ( 2u²cos²θ ) = 8tan40 – ( 9.8 * 8 * 8 ) / ( 2 * 20 * 20 * cos40 * cos40 ) = 5.4 m 

26 ) Prove that a gun will shoot three times as high when its angle of elevation is 60 degree as when it is 30 degree, but will carry the same horizontal distance.

case : 1 

initial velocity = u

angle of elevation ( Ф ) = 60 degree

height ( H ) = u²sin²Ф / g = u²sin²60 / g = 3u² / 4g

H = 3u² / 4g ———————————– ( 1 )

case : 2

initial velocity = u

angle of elevation ( θ ) = 30 degree

height ( h ) = u²sin²θ / g = u²sin30 / g = u² /4g

h = u² / 4g  ——————————- ( 2 )

( 1 ) ÷ ( 2 ) we get

H ÷ h = ( 3u² / 4g ) ÷ ( u² / 4g ) 

H / h = 3

H = 3h

Hence, a gun will shoot three times as high when its angle of elevation is 60 degree as when it is 30 degree, but will carry the same horizontal distance.

( Proved )

27 ) Prove that the maximum horizontal range is four times the maximum height attained by the projectile when fired at inclination so as to have the maximum horizontal range.

Solution : 

R = u² sin2Ф / g

where R = horizontal range

Ф = angle of projection 

u = initial velocity

g = acceleration due to gravity

for maximum R

sin2Ф = 1 

sin2Ф = sin90

2Ф = 90

Ф=45 

∴ R_max = u² / g  —————————————— ( 1 )

height ( H ) = u²sin²Ф2usinФ / g / 2g = u²sin²45 / 2g = u² /4g

H = u² /4g ——————————————————————–( 2 )

( 1 ) ÷ ( 2 ) we get 

R_max / H = ( u² / g ) ÷ ( u² /4g )              ( using equation (1 ) & ( 2 )  )

R_max = 4H 

( proved)

28 ) A bullet is fired with a velocity of 10 m/sec in a direction making an angle of 30 degree with the vertical. Calculate its time of flight and maximum height reached by it.

initial velocity ( u ) = 10 m /sec

angle made by bullet with vertical ( Ф ) = 30 degree

angle of projection ( θ ) = 90 – 30 = 60 degree

maximum height ( h ) = u² sin²θ/ 2g = (10² * sin²60 ) / (2*9.8) = 3.83 sec

time of flight ( T ) = 2usinθ / g = ( 2* 10 * sin60 ) / 9.8 = 10/9.8 = 1.77 sec 

29 ) A shell is fired with speed of 30 m/sec at an angle of 37 degree with the horizontal. After 1 second, the shell is moving at an angle θ with the horizontal. Calculate θ .  Given : sin37 = 3/5