Asked by Andres
A rock is tossed straight up with a velocity of 34.0 m/s. When it returns, it falls into a hole 17.3 m deep. What is the 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?
How long is the rock in the air, from the instant it is released until it hits the bottom of the hole?
Answers
Answered by
drwls
At the bottom of the hole, kinetic energy is increased by M g D from the original value. D is the depth.
Therefore V2^2/2 - V1^2.2 = g D
(The mass cancels out).
Solve for V2, which will be negative (going down)
V1 is the initial upward velocity
For the time in the air, T, solve
V1*T - (g/2)T^2 = -17.3 meters
Therefore V2^2/2 - V1^2.2 = g D
(The mass cancels out).
Solve for V2, which will be negative (going down)
V1 is the initial upward velocity
For the time in the air, T, solve
V1*T - (g/2)T^2 = -17.3 meters
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