Question

Two balls $A$ and $B$ are throw simultaneously, A vertically upwards with a speed of $20m/s$ from the ground and $B$ vertically downwards from a height of $40m$ with the same speed and along the same line of motion. At what point do the balls collide? $(g=9.8m/s^2)$

Answer 1

Suppose they meet at height 'h' from the ground in time 't'.

Initial velocity of projection be u=20 m/s

For the ball thrown upward (u and a are opposite)

$h=ut-\frac{1}{2}g{t}^2.............\left(i\right)$

For the ball thrown vertically downward (u and a are in same direction)

$40-h=ut+\frac{1}{2}g{t}^2........\left(ii\right)$

Add (i) and (ii) we get,

$40=2ut$

$t=\frac{40}{2u}=\frac{40}{2\times 20}=1sec$

Putting values in equation (i) we get,

$h=20\times 1-\frac{1}{2}\times 9.8\times 1^2$

$h=20-4.9$

$h=15.1m$

Balls will collide at height 15m from the ground in 1sec