# University Physics

A rock is tossed straight up with an initial speed of 20 m/s. When it returns, it fall into a hole that is 8.0 m deep. What is the velocity of the rock (in m/s) when it hits the bottom of the hole? Use -9.80 m/s2 for the local acceleration due to gravity and assume no significant air resistance. Assume that the y axis is positive in the up direction

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1. -8 = 0 + 20 t - 4.9 t^2

4.9 t^2 -20 t -8 = 0

v = 20 - 9.8 t

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2. I am sorry but I am still confused...

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3. more details:

h = original height + Vi t - (1/2)(9.8) t^2

-8 = 0 + 20 t - 4.9 t^2

4.9 t^2 -20 t -8 = 0

velocity = original velocity - 9.8 t
v = 20 - 9.8 t

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