A stone is thrown from the top of a cliff with the initial velocity of 20.0 m/s

Question

A stone is thrown from the top of a cliff with the initial velocity of 20.0 m/s straight upward. The cliff is 50.0 m high and the stone just misses the edge of the cliff on its way down. Determine (a) the time needed for the stone to reach its maximum height; (b) the maximum height, as measured from the ground; (c) the time needed for the stone to return to the level of the thrower; (d) the velocity of the stone at this instant; (e) the velocity and position of the stone at t=5.00 s, as measured from the ground; and (f) the velocity of the stone just before it hits the ground. A stone is thrown from the top of a cliff with the initial velocity of 20.0 m/s straight upward. The cliff is 50.0 m high and the stone just misses the edge of the cliff on its way down. Determine (a) the time needed for the stone to reach its maximum height; (b) the maximum height, as measured from the ground; (c) the time needed for the stone to return to the level of the thrower; (d) the velocity of the stone at this instant; (e) the velocity and position of the stone at t=5.00 s, as measured from the ground; and (f) the velocity of the stone just before it hits the ground.

Explanation / Answer

(A) v0 = 20 m/s

a = - 9.8 m/s^2

at max height, v = 0

vf = vi + a t

0 = 20 - 9.8 t

t = 2.04 sec

(B) vf^2 - vi^2 = 2 a d

0^2 - 20^2 = 2(-9.8)(y)

y = 20.4 m

from ground, Hmax = 50 + y = 70.4 m  

(C) for that, yf = yi

yf - yi = v0 t + a t^2 /2

0 = 20 t - 9.8 t^2 /2

t = 4.08 s

(D) v = vi + a t = 0+ (-9.8 x 4.08)

v = - 20 m/s  

(20 m/s downward )

(E) v = 20 + (-9.8 x 5) = - 29 m/s

(29 m/s downward)

y - 50 = (20 x 5) + (-9.8 x 5^2 /2)

y = 27.5 m

(F) v^2 - 20^2 = 2(-9.8)(0-50)

v = 37 m/s (downward) Or - 37 m/s

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