Asked by Cole
A stone was dropped off a cliff and hit the ground with a speed of 120 ft/s. Using the fact that acceleration due to gravity is −32ft/s2, what is the height (in feet) of the cliff?
Answers
Answered by
Reiny
a = -32
v = -32t + c
since it was 'dropped' , when t=0 , v =0
so c = 0
v = -32t
120 = -32t
t= 120/-32 = -15/4
s = -16t^2 + k
when it hit the ground, t = 15/4 , s = 0
0 = -16(225/16) = k
k = 225
s = -16t^2 + 225
So the height of the cliff is 225 ft
v = -32t + c
since it was 'dropped' , when t=0 , v =0
so c = 0
v = -32t
120 = -32t
t= 120/-32 = -15/4
s = -16t^2 + k
when it hit the ground, t = 15/4 , s = 0
0 = -16(225/16) = k
k = 225
s = -16t^2 + 225
So the height of the cliff is 225 ft
Answered by
Andrew
Minor note, although your final answer is correct, I would not that the time is not going to be negative because that final velocity is negative just like the acceleration, hence why the problem mentions the "speed" of the object. This ensures you have a positive time, because that is the only thing that makes sense in this situation logically. Good work though!
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