MOTION UNDER GRAVITY

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A BALL IS RELEASED FROM TOP OF TOWER OF HEIGHT H METERS. IT TAKES T SECONDS TO REACH THE GROUND. WHAT IS ITS POSITION OF THE BALL IN T/3 SEC???

Guest Jun 17, 2016

Guest · Answer 1 · 2016-06-17T11:59:26

Use this formula for your problem. H=1/2 g t^2, where H=Height, g=9.81 m/s^2 gravitational constant, t=time in seconds.

Sometimes "d" is used instead of "H", which stands for "distance".

Alan · Answer 2 · 2016-06-17T12:05:06

Since the height of the tower is given as H in this instance, the final position here is given by:

H - 1/2g(T/3)^2 meters above the ground (where g = 9.81 m/s^2)