1 answer

Velocity, time, position

Question:

A stone is thrown from the top of a building with initial veocity of 20 m/s upwards. The building is 50 m. high and the stone just misses the edge of the roof on itsway down. Determine:
(a) The time needed for the stone to reach its maximum height?
(b) The position and velocity of the stone at t=5 seconds?

Answers

Initial Speed of the stone u = 20 m/s
Height of the building H = 50 m
(a)
Time needed to reach maximum height ta = u/g = 20/9.8 =2.04 s
(b)
Velocity at t = 5s, v = 9.8 * (5 - 2.04) = 9.8 * 2.96 = 29.008 m/s
Time taken to reach the point which is at the same level to the top of the
building t' = 2ta = 4.08 s
Position of the stone at t = 5s, = (1/2)(9.8)(5 - 4.08) = 4.9*0.92 = 4.508 m from the top of the building.
Position of the stone from ground, Y = 50 - 4.508 = 45.492 m



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