physics help! Problem 2.51 MC Part A A rock climber stands on top of a 51 m -hig
ID: 1652434 • Letter: P
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physics help!
Problem 2.51 MC Part A A rock climber stands on top of a 51 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 2.5 m/s. In all parts, use the standard sign convention for direction How long does it take the first stone to hit the water? Express your answer using three significant figures and include the appropriate units Hints | t 1= 1 Value Units Submit My Answers Give Up Part B What was the initial velocity of the second stone? Express your answer using three significant figures and include the appropriate units. Use the standard sign conventions for direction. Hints Ui,2 %2 1 Value Units Submit My Answers Give UpExplanation / 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 - 2t +51
t=3 (time after first stone's release until both stone hit the water)
so it only took the second stone 2s to hit the water after its own release since 3-1=2
0 = -4.9 x 4 + 2v + 51
v = -15.7 (initial velocity of second stone)
For c part just multiply the respective times by acceleration and add to the initial velocities of the stones.
So for the first stone
v(final) = -2 - 9.8x3 = -31.4 m/s
and for the second stone
v(final) = -15.7 - 9.8x2 = -35.3 m/s
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