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An object is dropped from rest from a 70.6 m tower. Air resistance is negligible. After 0.32 seconds, what is magnitude and direction of its acceleration?

Sagot :

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Answer:

1,378.9ms²

Explanation:

Given the following

Distance S = 70.6m

Time t = 0.32secs

Initial velocity = 0m/s

Required

Acceleration

Using the equation of motion

S = ut+1/2at²

Substitute

70.6 = 0+1/2a(0.32)²

70.6 = 0.0512a

a = 70.6/0.0512

a = 1,378.9

Hence the acceleration is 1,378.9ms²

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