physics
posted by Amanda .
A rock is tossed straight up with a speed of 29m/s. When it returns, it falls into a hole 13m deep.
Solve for rock's velocity as it hits the bottom of the hole.
&How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?

(1/2)M V2^2 = (1/2)MV1^2 + M g D
D is the depth of the hole.
V1 is the initial "toss" velocity
V2 is the velocity at the bottom of the hole.
Cancel out the M's and solve for V2.
For time in the air, solve the following equation for t:
Y = 13 = V1*t  (g/2)*t^2
g = 9.8 m/s^2