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Fynjy0 [20]
3 years ago
14

If you were to drop a rock from a tall building, assuming that it had not yet hit the ground, and neglecting air resistance, aft

er it has fallen 22 m:
How much time has passed (in s)?
what is the speed at this distance (in m/s)?
Physics
1 answer:
elena-14-01-66 [18.8K]3 years ago
5 0

Since the object is dropped from some height so its initial speed must be zero

acceleration of the object is due to gravity

so we can use kinematics to find the time it will take to drop by x = 22 m

\delta x = v_i * t + \frac{1}{2}at^2

22 = 0 + \frac{1}{2}*9.8*t^2

t = 2.12 s

Now the speed after 2.12 s will be given as

v_f = v_i + at

v_f = 0 + 9.8 * 2.12

v_f = 20.8 m/s

so above is the speed and time

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Consider a simple pendulum with a period
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1. The length is 8.35m

2. The period on the moon is 14.05 secs

Explanation:

1. Data obtained from the question. This includes the following:

Period (T) = 5.8 secs

Acceleration due to gravity (g) = 9.8 m/s2

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The length can be obtained by using the formula given below:

T = 2π√(L/g)

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Take the square of both side

(5.8)^2 = 4π^2 x L/ 9.8

Cross multiply

4π^2 x L = (5.8)^2 x 9.8

Divide both side by 4π^2

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Length (L) = 8.35m (the length remains the same)

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The period can be obtained as follow:

T = 2π√(L/g)

T = 2π√(8.35/1.67)

T = 14.05 secs

Therefore, the period on the moon is 14.05 secs

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