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3 years ago

A ball initially at rest rolls down a hill with an acceleration of 3.3 m/s2. If it accelerates for 7.5 s, how far will it move?

Physics

1 answer:

goldfiish [28.3K]3 years ago

5 0

<u>Answer:</u>

Ball will move 92.8125 meter along the cliff in 7.5 seconds.

<u>Explanation:</u>

We have equation of motion , $s= ut+\frac{1}{2} at^2$ , s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

In this case initial velocity = 0 m/s, acceleration = 3.3 $m/s^2$ , we need to calculate displacement when time = 7.5 seconds.

Substituting

$s=0*7.5+\frac{1}{2} *3.3*7.5^2\\ \\ =92.8125 meter$

So ball will move 92.8125 meter along the cliff in 7.5 seconds.

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