A rock is tossed straight up with a velocity of 33.1m/s. When it returns, it falls into a hole 17.2m deep. What is the rock's velocity as it hits the bottom of the hole? 

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A rock is tossed straight up with a velocity of 33.1m/s. When it returns, it falls into a hole 17.2m deep.

The velocity of the rock when it returns to the same level it was thrown up from is the same as its initial velocity. As the rock was thrown up with a velocity of 33.1 m/s, at ground level it has a velocity of 33.1 m/s.

Moving lower into the hole 17.2 m deep the velocity of the rock increases and reaches V at the bottom of the hole.

2*9.8*17.2 = V^2 - 33.1^2

=> V^2 = 33.1^2 + 2*9.8*17.2

=> V^2 = 11432.73

=> V = `sqrt 11432.73`

=> V = 37.85

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