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 ?

  y/y₀ = 1/2 + v₀²/(2 g y₀)

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  

= - 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)

the force required to move the cart is divided by three or 33.33 n.

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