A mountain climber stands at the top of a 50.0 m cliff hanging over a calm pool

Question

A mountain climber stands at the top of a 50.0 m cliff hanging over a calm pool of water. The climber throws two stones vertically 1.0's apart and observes that they cause a single splash when they hit the water. The first stone has an initial velocity of + 2.0m/s. How long after release of the first stone will the two stones hit the water? What is the initial velocity of the second stone when it is thrown? What will the velocity of each stone be at the instant both stones hit the water? A model rocket is

Explanation / Answer

consider upward direction to be positive -s = ut-0.5g*t^2 -50 = 2t -4.9t^2 solving t we get t=3.4s let us assume v is in downward direction -v=u-gt so v=-u+gt = -2+(9.8*3.4) = 31.32m/s both stones hit after 3.4s from first stone release after 1sec,second stone is thrown t=2.4s so -50 = u(2.4) -4.9*(2.4)^2 u= -9.1m/s intial velocity of second stone = 9.1m/s vertically downward (-ve sign) -v= u-gt so v=-u+gt = 9.1+(9.8*2.4)=32.62m/s

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