please help A workman (accidentally) drops a wrench over the side of a high-rise

Question

please help

A workman (accidentally) drops a wrench over the side of a high-rise building. Someone looking out of a window on a floor below sees the wrench appear (falling past them) at exactly 11. 00 am. Someone in an office 30. 0 m below the first observer sees the wrench appear (as it continues to fall) 1. 429 seconds later. Calculate the average velocity of the wrench during the 1. 429 second interval. Calculate the instantaneous velocity of the wrench when it passes the second observer. Calculate the vertical distance the wrench has fallen from the point at which it was released when it passes the SECOND observer.

Explanation / Answer

Let u and v be the velocities at first and second observer.

s = 30 = ut + gt^2/2 = 1.429u + 9.8*1.429^2/2 => u = 13.99 m/s

v = u + gt = 13.99 + 9.8*1.429 = 27.99 m/s

Let h be the distance to the second observer from point of release.

v^2 - 0^2 = 2gh => h = v^2/2g = 37.97m

Answers are:

a) average velocty = distance/time = 30/1.429 = 20.99m

b) v = 27.99 m/s

c) h = 37.97 m

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