Now a stone is thrown upward at the top of a cliff that is 102 meters high. It i

Question

Now a stone is thrown upward at the top of a cliff that is 102 meters high. It is thrown upward at a speed of 5.00 meter/second. (a) At the top of its path its velocity is momentarily zero. Use this information to find out how much time it took for the stone to get there after it was released. (b) How high is this highest point above the ground? (c) When will it reach the ground: find the time between throwing it and reaching the ground. (d) What is the speed of the stone just before it hits the ground?

Explanation / Answer

(A) vf = 0

vi = 5 m /d

a = - g = - 9.8 m/s^2

vf = vi + a t

0 = 5 - 9.8t

t = 0.51 sec

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

0^2 - 5^2 = 2 (-9.8) h

h = 1.28 m

H = 102 + h = 103.28 m

(C) to reach ground,

vi = 0

y = -103.28 m

y = v0 t + a t^2 / 2

-103.28 = 0 - 9.8 t^2 /2

t = 4.59 sec

total time = 4.59 + 0.51 = 5.10 sec

(d) vi = 5 m/s

t = 5.10 s

v = 5 - (9.8 x 5.10) = - 45m/s

speed = 45 m/s downward

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