Question

Starting from rest, a child throws a ball with an initial speed v at B degrees above the horizontal. The child then chases after the ball, accelerating at a constant acceleration a. If the child wants to catch the ball at the same height it was thrown, what must be the child's acceleration a? Use g for the gravitational constant.

Answer 1

Let the child accelerate with 'a' to catch the ball after travelling some distance 's'. let t be the time taken to catch the ball

From equation of motion for this horizontal motion,

s= 1/2 a t^2

for horizontal component of the balls motion

s= vCosB.t

So,

1/2 a t^2 = vCosB.t .... Ist equation

For verticle projectile motion of ball,
At highest point of the ball, let time taken be T

0=vSinB-gT

T=vSinB/g

Since the ball takes twice the amount of time to reach the highest point,

t= 2vsinB/g .... II nd Equation

from I and II

1/2 a (2vSinB)/g = vCosB

a = gCosB/SinB = gCotB

Hence a=gCotB