An elevator is descending vertically at a constant speed of 10.0 m/s. A loose bo

Question

An elevator is descending vertically at a constant speed of 10.0 m/s. A loose bolt falls from the bottom of the elevator car at t = 0 when the elevator car is 75.0 m above the bottom of the elevator shaft. How long does it take for the bolt to hit the bottom of the elevator shaft? What is the velocity of the bolt when it hits the bottom of the elevator shaft? The elevator continues to travel downwards at a constant speed of 10.0 m/s until it is 50.0 m from the bottom of the shaft, then decelerates at a uniform rate so that it just comes to rest as it reaches the bottom of the shaft. How long does the elevator take to travel the 75.0 m to the bottom of the elevator shaft?

Explanation / Answer

initial speed=10

acceleration=9.8 m/s^2

height=75

so if time is t,

then 10*t+0.5*9.8*t^2=75

t=3.02 sec

b)velocity=10+9.8*3.02=39.596 m/s

c)time taken to travel 50 m=50/10= 5 sec

then time taken to travel the rest 25 m let be t.

then if deceleration is a,

0^2-10^2=2*a*25

a=-2 m/s^2

so 25=10*t-0.5*2*t^2

t=5 sec

so total time taken=10 sec

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