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A ball is launched straight up with initial speed of 30.0 m/s. What is the ball's velocity when it comes back to its original point

Sagot :

Answer:

We could get the time taken by the ball to return back to earth, using the formula:

s = u t + ½ a t², where

s = displacement of the body moving with initial velocity u, acceleration 'a' in time t.

In the present case s=0 (as the ball returns back to starting time)

u= 30 m/s; a = -10 m/s² ( negative sign as a is in opposite direction to u); t=?

0 = 30 t - ½ ×10 ×t²; ==> 5 t = 30, t= 6 second.

So ball will return back after 6 second after being thrown up.

Explanation:

I looked it up

Hope this helps

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