A freely falling object requires 1.35s to travel the last 28.5 m before it hitst

Question

A freely falling object requires 1.35s to travel the last 28.5 m before it hitsthe ground. From what height above the ground did it fall?
I do not understand how you are supposed to find the totaldistance when your are just given the end of the objectfalling. A freely falling object requires 1.35s to travel the last 28.5 m before it hitsthe ground. From what height above the ground did it fall?
I do not understand how you are supposed to find the totaldistance when your are just given the end of the objectfalling.

Explanation / Answer

Since d=vi*t+1/2a*t^2, 28.5=vi*1.35+1/2*9.8*(1.35)^2, solvingfor vi, you get vi=14.5m/s.
Since the object was not stated to have fallen with anyinitial velocity, we assume that it fell from rest, thus vi=0 so, since we got the velocity when the 1.35 seconds happens,we make that the final velocity and use the formula vf^2=vi^2+2ad to find the d. thusd=(vf^2-vi^2)/(2a)=(14.5^2-0^2)/(2*9.8)=10.73 m.
But we still need to add the last 28.5 m to this so we get28.5+10.73=39.2 m, which is the height the object fell.

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