A rock climber stands on top of a 57 m -high cliff overhanging a pool of water.

Question

A rock climber stands on top of a 57 m -high cliff overhanging a pool of water. He throws two stones vertically downward 1.0 s apart and observes that they cause a single splash. The initial speed of the first stone was 1.5 m/s, In all parts, use the standard sign convention for direction. What is the velocity of the first stone as it hits the water? Express your answer using three significant figures and include the appropriate units. Use direction the standard sign direction. What is the velocity of the second stone as it hits the water? Express your answer using three significant figures and include the appropriate units. Use the standard sign direction.

Explanation / Answer

Well since they made a single splash, they hit the water at the same time and therefore the time after the release of the first stone until the second stone hits the water is also the time it takes the first stone to hit the water after the first stone's release.
0 = -4.9t^2 - 1.5t +57
t=3.26 (time after first stone's release until both stone hit the water)
so it only took the second stone 2.26 s to hit the water after its own release since 3.26-1=2.26
0 = -4.9 x (2.26)^2 + 2.26v + 57
v = -14.14 (initial velocity of second stone)

So for the first stone
v(final) = -1.5 - 9.8x3.26 = -33.45 m/s
and for the second stone
v(final) = -14.14 - 9.8x2.26 = -36.28 m/s

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