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Olin [163]
2 years ago
11

A rock is thrown downward from the top of a cliff with an initial speed of 20 m/s. If the rock hits the ground after 5.0 s, what

is the height of the cliff? (Disregard air resistance. a =-g=-9.81 m/s2.)​
(i want step by step)
Physics
1 answer:
Vinil7 [7]2 years ago
7 0

Answer

I hope it's helps you

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Answer:

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Explanation:

given data

distance = 53.2 m

height = H

to find out

height H

solution

we know height is here H = \frac{1}{2} gt^2    ......................1

here t is time and a is acceleration

so

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we know during time (t -1) s , it fall distance (H - 53.2) m

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H - 53.2 = \frac{1}{2} g (t-1)^2

H - 53.2 = \frac{1}{2} g (t^2-2t+1)

H - 53.2 = \frac{1}{2} gt^2-gt+\frac{1}{2} g     ................2

now subtract equation 2 from equation 1 so we get

H - (H - 53.2) =\frac{1}{2} gt^2- (\frac{1}{2} gt^2-gt+\frac{1}{2} g)

53.2 = gt - \frac{1}{2} g

53.2 = 9.81 t - \frac{1}{2} 9.8

t = 5.92 s

so from equation 1

H = \frac{1}{2} (9.81)5.92^2

H = 171.90 m

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