Ball A is dropped from the top of a building of height h at the same instant tha

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 n opposite directions, and the speed of A is twice the speed of B. At what height does the collision occur? (Use any variable or symbol stated above as necessary.) Both balls experience constant acceleration once they are n flight. Choose a coordinate system with the origin at the ground and the upward direction positive. When the balls collide they are at the same height above the ground.

Explanation / Answer

from the kinmeatic equation

the distance of ball A is

xA = v0 t + (1/2) at^2

v0 t = velocity ( time ) = distance

since ball is at height the above equation changes as

xA = H - ( 1/2) gt^2

for ball B

xB = v0 t -( 1/2) gt^2

the condition of collision is

xA = xB

vA =- 2vB

from the kinematic equation

the speed of the ball A is

vA = u- gt

since initial speed of the ball A is zero

vA = -gt

the speed of the ball B is

vB = v0 - gt

since vA = - 2vB

    -gt =- 2 ( v0 - gt)

-gt = -2 v0 - 2gt

3gt =2 v0

t = 2v0/3g

since xA = xB

H - ( 1/2) gt^2=v0 t -( 1/2) gt^2

H = v0 t

= v0 (2v0/3g)

= 2 v0^2/ 3g

xA = 2 v0^2/ 3g- ( 1/2) gt^2

   = 4 v0^2/9g = (2/3) H

so, (2/3) times height collision occur

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