A stone is dropped from the top of a cliff and is found to ravel 44.1m diving the last second before it reaches the ground. What is the height of the cliff?
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s = 44.1m
a = 9.81 m/s
t = 1s
Use s = ut + 0.5at^2.
s = distance, u = initial velocity (1 second before impact), t = time, a = acceleration
44.1 = u + 0.5*9.81*1
44.1 = u + 4.905
u = 44.1 - 4.905 = 39.195 m/s.
Use v^2 = u^2 + 2as where v = final velocity = 39.195m/s (1second before impact)
u = 0, a = 9.81m/s s = distance
39.195^2 = 0 + 2*9.81s
s = 39.195^2/(2*9.81)
s = 78.3 m
Total distance ( = height of cliff) = 78.3 + 44.1 = 122.4 m = Answer
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