A BALL IS RELEASED FROM A HEIGHT OF 45m TOP OF THE BUILDING. IF IT STRIKES THE GROUND AFTER 5 SECOND, CALCULATE THE ACCELERATION AND FINAL VELOCITY OF THAT BALL
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Explanation:
apply equation of motion
S= ut+ 1/2 at^2
here s = 45m
t=5 sec.
than in question given:- ball is released so U= 0
than after solving you will get the value of a= 18/5 m/s^2.
than after that , again apply another equation of motion
V= u+ at
= 0+ 18/5× 5
= 18m/s .
I hope you will get your answer........
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