A mountain climber stands at the top of a 53.4m cliff that overhangs a calm pool of water. He throws two stones vertically downward 1.79s apart
He throws two stones vertically downward 1.79s apart and observes that they cause a single splash. The first stone has an initial velocity of 1.84 m/s.
What initial velocity must the second stone have if they are to hit simultaneously?
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The climber throws two stones vertically downwards from the top of a cliff that is 53.4 m high. The second stone is thrown 1.79 s after the fist stone. If the stones cause a single splash, the distance of 53.4 m is traveled by the second stone in a time period that is 1.79 s less than the first. This can be done by changing the initial velocity at which it is thrown. The first stone has an initial velocity of 1.84 m/s. The time taken by it to strike the water is T, where 53.4 = 1.84*T + (1/2)*9.8*T^2
=> 1.84T + 4.9T^2 = 53.4
Solving the equation gives the positive root of 3.1187 s
If the 2nd stone is thrown with a velocity V, V(3.1187-1.79) + 4.9*(3.1187 - 1.79)^2 = 53.4
=> V = 33.67 m/s
The second stone has to be thrown with a velocity 33.67 m/s for it to strike the water with the first stone.
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