A mountain climber stands at the top of a 45.0 m cliff that overhangs a calm poo

Question

A mountain climber stands at the top of a 45.0 m cliff that overhangs a calm pool of water. She throws two stones vertically downward 1.00 s apart and observes that they cause a single splash. The first stone had an initial velocity of -3.00 m/s.
(a) How long after release of the first stone did the two stones hit the water?
(seconds)
(b) What initial velocity must the second stone have had, given that they hit the water simultaneously?
(m/s)
(c) What was the velocity of each stone at the instant it hit the water? first stone (m/s)
second stone-(m/s)

Explanation / Answer

dist. = v * t + 1/2 * g * t^2 describes the motion of a stone with initial speed. Replacing with numerical values 56.8 = 2.37 * t + 1/2 9.8 m/s * t^2 this is a good old quadratic: 2.37 * t + 1/2 9.8 m/s * t ^2 - 56.8 = 0 t = 3.1714 seconds is the time for the first stone. Now the second stone is late. it has to be there in 3.17 - 1.38 = 1.79 s again with dist. = v * t + 1/2 * g * t^2, but this time we do not know v (56.8 - 0. 5 * 9.8 * 1.79 ^2) / 1.79 = 22.96 m / s as initial speed the second stone has a final speed of: v + a * t = 22.96 + 9.8 * 1.79 = 40.5 m/sec or 145 km/h

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