Question

Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions, and the speed of A is twice the speed of B. At what height does the collision occur?

Answer 1

Answer:

(2/3) times height collision occur

Explanation:

for ball A

from the kinematic equation

the distance of ball A is

$x_A = v_0 t + \frac{1}{2} at^2$

$v_0* t$= velocity ( time ) = distance

since ball is at height, the above equation changes as

$x_A = H - \frac{1}{2} gt^2$

for ball B

$xB = v0 t - \frac{1}{2} gt^2$

the condition of collision is

xA = xB

$vA = - 2vB$ (given)

from the kinematic equation

the speed of the ball A is

$v_A = u- gt$

since initial speed of the ball A is zero

, so

$v_A = -gt$

the speed of the ball B is

$v_B = v_0 - gt$

since$v_A = - 2v_B$

   $-gt = -2 ( v_0 - gt)$

$-gt = -2 v_0 +2gt$

$3gt =2 v_0$

$t = \frac{2v_0}{3g}$

since $x_A = x_B$

$H - \frac{1}{2} gt^2= v_0 t - \frac{1}{2} gt^2$

$H = v_0 t$

$= v_0 (2v_0/3g)$

$= \frac{2 v_0^2}{ 3g}$

$x_A = 2 (\frac{v_0^2}{ 3g})- \frac{1}{2} gt^2$

  $= 4 \frac{v_0^2}{9g} = (2/3) H$

so, (2/3) times height collision occur