Problems on oblique projectile

Q1

Prove that the velocity at the end of flight of an oblique projectile is the same in magnitude as at the beginning but the angle that it makes the horizontal is negative of the angle of projection.

Sol:

Let u be the initial velocity with which the object is projected at angle 'θ' with the horizontal

Let V be the velocity at the end of flight.

Horizontal component of v = v_x = u cos θ,

This is because, horizontal motion is a uniform motion

Vertical component v_y = u_y + a_yt
v_y = u sin θ – gT

T is the time of flight

∴ v_y = u sin θ – 2u sin θ

v_y = – u sin θ

The resultant velocity

v = &sqrt;u²(sin²θ + cos²θ = 1)

∴ v = u

So, the magnitude of the velocity at the end of the flight is equal to the initial velocity

Now the direction of the final velocity

is

∴ The angle made by the velocity vector at the end of the flight is negative of the angle of projection

Q2

A football player kicks a foot ball at an angle θ = 40° above the horizontal axis with initial velocity 22m/s. Ignore the air resistance and find the maximum height that the ball attains, time of flight between kickoff and landing and the range of the ball ?

Sol:

Given, Initial velocity of the ball u = 22m/s

Angle of projection θ = 40°

Acceleration due to gravity g = 9.8m/s²

Maximum height, μ

Time of flight, T

Horizontal range, R

Q3

A football is kicked off with an initial speed of 19.6 m/s to have maximum range. Goal keeper standing on the goal line 67.4 m away in the direction of the kick starts running opposite to the direction of kick to meet the ball at that instant. What must his speed be if he is to catch the ball before it hits the ground?

Sol:

or R = 39.2 metre.

Man must run 67.4m – 39.2m = 28.2m in the time taken by the ball to come to ground time taken by the ball.

t	=	2&sqrt;2
	=	2 × 1.41
	=	2.82 sec
Velocity of man	=
	=	10 m/sec

Q4

In the absence of wind the range and maximum height of a projectile were 'R' and 'H'. If wind imparts a horizontal acceleration a =(g/4) to the projectile then find the maximum range and maximum height.

Sol:

H¹	=	H (∴ u sinθ remains same)
T¹	=	T
R¹	=	u_x + (1/2)aT²
	=	R + (1/2)(g/4)T²
	=	R + (1/8)gT²
	=	R + H
R¹	=	R + H
H¹	=	H

Q5

A body projected from a point 'O' at an angle θ, just crosses a wall 'y' m high at a distance 'x' m from the point of projection and strikes the ground at 'Q' beyond the wall as shown, then find height of the wall.

Sol:

We know that the equation of the trajectory is

y	=	x tanθ – can be written as
y	=	x tanθ –
&doublearrow; y	=	x tanθ –
y	=	x tanθ –
y	=	x tanθ