Example 3 A stone is thrown from the top of a 52.5-m high vertical cliff and lands in the water below at a location 43.8 m from the bottom of the cliff. Determine the velocity with which the stone is thrown.
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Answer:
The acceleration due to gravity is -9.8 m/s^2, since it acts downwards.
Using the equation of motion for vertical motion, we have:
h = v*t + (1/2)at^2
where h is the initial height of the stone (52.5 m), t is the time it takes for the stone to reach the water, and we have substituted a = -9.8 m/s^2.
Solving for t, we get:
t = sqrt((2h)/a) = sqrt((252.5)/9.8) = 3.19 s
Now, using the equation of motion for horizontal motion, we have:
d = v*t
where d is the horizontal distance traveled by the stone (43.8 m).
Solving for v, we get:
v = d/t = 43.8/3.19 = 13.74 m/s
Therefore, the velocity with which the stone is thrown is 13.74 m/s.