A stone is thrown from the top of a building with an initial velocity of 23.3 m/

Question

A stone is thrown from the top of a building with an initial velocity of 23.3 m/s straight upward, at an initial height of 55.0 m above the ground. The stone just misses the edge of the roof on its way down

a. Determine the time needed for the stone to reach its maximum height.

b. Determine the maximum height.

c. Determine the time needed for the stone to return to the height from which it was thrown, and the velocity of the stone at that instant.
time:_____
Velocity:_____

d. Determine the time needed for the stone to reach the ground.

e. Determine the velocity and position of the stone at t = 5.81 s.
Velocity:_____ m/s
Position:_____ m

Explanation / Answer

Let the time taken be t At the maximum height its velocity,v = 0/sec v = u + at Thus 0 = 23.3 - 9.8*t Thus t = 2.4 sec Let the Maximum height be h thus v^2 - u^2 = 2as Thus s = u^2/2g = 27.7 m The time to return to the same level is 2 times of the time it takes to attain maximum height Thus T = 2t = 4.8 sec Velocity will be same in magnitude and opposite in direction Velocity = - 23.3 m/sec

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