Posted by Elisabeth on Thursday, July 8, 2010 at 10:32pm.
You have the right formula.
Let S0 be the height of the cliff in meters.
When the stone stops it is at 0m height.
So,
0m = S0 + (v0)t + (1/2)gt^2
t =7 s
g=-9.8m/(s^2)
0 = S0 + (20m/s)(7s) + (1/2)(-9.8m/(s^2))(49s^2)
0 = S0 + 140m - 240.1m
S0 = 100.1m
Assuming that from the bottom to the top of the cliff is h1 and the top of cliff to the top of the parabola is h2... isn't the 100.1m really h1 + h2 (since the stone is thrown upwards)? I am looking for just h1.
In the equation
s(t) = (s0) + (v0)t + (1/2)(a)(t^2)
s(t) gives the height for some time t.
s0 is the initial height, which is the height of the cliff.
v0 is the initial velocity, for this problem 20m/s. The last term is the distance due to gravity. As t^2 increases, this term finally overwhelms the others and the stone reverses direction and falls back to a point at the base of the cliff.
How high the stone travels above the cliff can be calculated, but it is not needed to find the height of the cliff.
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