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

Question

4. 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) at the top of speed v= 0

u = 5m/s

v = u - at

0=5 - 9.8 *t

t=5/9.8 s = 0.51 s

b) v2 = u2 + 2aS

25 = 0 + 2*9.8*s

s =1.28m

now total height = height of cliff + 1.28m = 103.28 m

c) throwing up to top time = 0.51 s

now coming from top to bottom total height = 103.28 m

initial velocity now u = 0

coming down with we dont know v

so using v2 = u2 + 2aS

= 0 + 2x9.8x103.18

= 2022.328

v = 44.97 m/s

now using v = u + at

44.97 = 0 + 9.8 *t

t= 4.58 s

total time = 4.58 +  0.51 s = 5.0988 s

d) v = u+ at

u = 0

v= at

= 9.8 *4.58 = 44.97 m/s

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