An elevator repairman is working at the top of an elevator shaft 60.0 m above th

Question

An elevator repairman is working at the top of an elevator shaft 60.0 m above the top of the elevator car. He drops a coin just as the car begins to move upwards. Th elevator first accelerates from rest at a constant rate equal to one tenth the acceleration due to gravity for 3 s, then continues upward at a constant speed.
(a) How long does it take the coin to hit the top of the car?
(b) If the speed of sound is 340 m/s, how long does it take the repairman to hear the coin hit the car.
(For an optional challenge consider if the coin bounces back up with 10% of its impact speed while the car continues upward at its contstant rate, how long until the repairman hears the second bounce?)

Explanation / Answer

a) suppode after tym t:

0 + 0.98 x 3^2/2 + 0.98 x 3 x t   + 9.8 x 3^2/2 + 9.8 x 3 x t + 9.8 x t^2/2 = 60

4.9t + 32.34t -11.49 = 0

solving quadratic etn.

t = 0.339

T = 3 + 0.339 = 3.34 sec

b) d = 9.8 x 3.34^2/2 =54.66 m

t1 = 54.666/340 = 0.161 sec

T = 3.34 + t1 =3.50 sec

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