Ball A is dropped from top of a building of height \"H\" at the same instant Bal

Question

Ball A is dropped from top of a building of height "H" at the same instant Ball B is shot vertically upward from the ground. When the balls collide they are moving in oppsite directions and the speed of A is twice the speed of B.
Given : H. Determine the heigh of the collision point.


A car starts from rest and accelerates at "a" for "T" secomds. The speed is then hold constant for another "T" seconds. The car then de-accelerates at "a/2" until it stops.
Given: a, T
Determine: The total distance traveled by car

Explanation / Answer

let ball A travelled distance h1 and ball B h2 before they meet .
h1+ h2 = H

Let after time "t" balls collide.
for ball A,
as it is dropped from the top, it's initial velocity is 0
Applying v=u+at

v=0+gt

v=gt ........(1)

for ball B,

Applying v=u+at

v/2 = u - gt ..............(2) (since speed of ball B is half of that of ball A )

from (1) and (2)

we get u=3gt/2

Now for ball A

H-h = 1/2 gt2 .............(3)

for ball B

h=(3gt/2)t - 1/2 gt2

h= gt2 ...............(4)

from (3) and (4)

h=2H/3

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