Question

Ball A is dropped from the top of a building. At the same instant ball B is thrown vertically upwards from the ground. When the balls collide, they are moving in opposite direction and the speed of A is twice the speed of B. At what fraction of the height of the building did the collision occurs ?

Answer 1

y/y₀ = 1/2 + v₀²/(2 g y₀)

Step-by-step explanation:

This in a kinematics exercise in a mention

ball A.

Since the ball is dropped, its velocity starts at zero, at the meeting point the equation is

`v_(A)^2`= - 2 g (y₀-y)

ball B

v_{B}^2 = v₀² - 2 g y

we substitute

2v_{B}^2 = -2 g (y₀ -y)

v_{B}^2 = - g y₀ + 2g y

v_{B}^2 = v₀² - 2gy

we have a system of two equations with two unknowns, therefore it can be solved. Let's multiply-by -1 and add

0 = g y₀ + v₀² -2gy

we clear the height

y = (g yo + v₀²) / 2g

y = yo / 2 + v₀² / 2g

In this exercise we assume that the height of the building is known and the initial velocity of ball B

The fraction is

y/yo = 1/2 + v₀²/(2gyo)