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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