A mountain climber stands at the top of a 57 m cliff that overhangs a calm pool

Question

A mountain climber stands at the top of a
57 m cliff that overhangs a calm pool of water.
He throws two stones vertically downward
1.15 s apart and observes that they cause
a single splash. The first stone has an initial
velocity of -2.15 m/s.

How long after release of the first stone will
the two stones hit the water? The acceleration
of gravity is 9.8 m/s2 .
Answer in units of s

What initial velocity must the second stone
have if they are to hit simultaneously?
Answer in units of m/s

What will be the velocity of the second stone
the instant both stones hit the water?
Answer in units of 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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