A bolt comes loose from underneath an elevator that is moving upward at a speed

Question

A bolt comes loose from underneath an elevator that is moving upward at a speed of 5.3 m/s. The bolt reaches the bottom of the elevator shaft in 3.9 s. (a) How high up was the elevator when the bolt came loose? 2.73 (b) What is the speed of the bolt when it hits the bottom of the shaft? m/s In the absence of air resistance, the brick experiences constant acceleration and you can use constant-acceleration equations to describe its motion. Be careful when assigning signs to the kinematic quantities. eBook

Explanation / Answer

Height gain after coming loose = (v^2/2g) = 1.433 m

Time to come to stop = (v/g) = 0.5408 sec.

Time available to drop t = (3.9 - 0.5408) = 3.36 secs.

Drop distance h = 1/2 (t^2 x g) = 55.29 metres.

a) Height when bolt parted = (55.29 – 1.433) = 53.86 m

b) V at bottom = sqrt (2gh) = sqrt (2*9.8* 55.29) = 32.92 m/sec.

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