what angle should a body be projected with a velocity

At what angle should a body be projected with a velocity 24 ms^-1 just to pass over the obstacle 16 m high at a horizontal distance of 32 m . Assume g = 10 ms^-1?

Solution:

what angle should

If point of projection is taken as the origin of the coordinate system, the projected body must pass through a point having coordinates (32 m, 16 m) as the following figure.



Suppose u be the initial velocity of the projectile. If θ be the angle of projection, then

Horizontal component of initial velocity u_x = u cos θ.

And vertical component of initial velocity u_y = u sin θ.

Suppose the body passes through point P at time t. then horizontal distance covered is x = 32 m i.e., 32 = x = (u cos θ ) t or 32 = (24 cos θ) t …….(1)

Similarly, vertical distance covered is y = 16 m i.e.,


16 = y = (u sin θ)t-1/2gt²

Or  16 = (24 sin θ)t – 1/2x 10x t² …..(2)

From equation (1) we get : t = 32/(24 cos θ)

Putting this value of t in equation (2), we have:

16 = (24 sin θ) 32/(24 cos θ) – ½ x 10x (32/24cos θ)²

Or 16 = 32 tan θ – 5 x 16/9cos² θ

Or 1 = 2 tan θ – 5/9 sec² θ or 9 = 18 tan θ – 5(1+tan² θ)

Or 5 tan² θ – 18 tan θ + 14=0

Solving above equation: tan θ = [18 ±√(18)² -4 x 5 x 14]/10= 2.462 or 1.137

So, Θ = 67⁰54’ or 48⁰40’