Posted by **TechnoBoi11** on Thursday, January 25, 2007 at 4:14pm.

A ball is thrown vertically downward from the top of a 35.9-m tall building. The ball then passes the top of a window that is 10.7 m above the ground 2.00 s after being thrown.

What is the speed of the ball as it passes the top of the window?

The equation for the speed of the ball, from time t=0 when the ball is released, is

V = Vo + g t

Vo it the initial velocity of the ball. The distance it has travelled at time t, measured vertically downward, is

Y = Vo t + (1/2) g t^2

You know that Y = 25.2 m at t = 2.

Use that information and the last equation to determine Vo. Then use that Vo in the first equation to get the value of V when t = 2.

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