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What is the time taken by a ball thrown up at 45 m/s to return to where it was thrown...

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xetatheta | Student, Kindergarten | Salutatorian

Posted May 20, 2012 at 2:48 PM via web

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What is the time taken by a ball thrown up at 45 m/s to return to where it was thrown from?

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted May 20, 2012 at 2:56 PM (Answer #1)

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A ball is thrown upwards at 45 m/s. The ball initially rises but due to the acceleration of gravity in the reverse direction its upward speed decreases constantly till it reaches a value of 0. The ball then stops and begins to fall downwards returning to where it had started from. The time taken by the ball to rise is equal to the time taken by it to fall.

Determine the time taken by the ball to reaches the highest point using the formula v = u + a*t. v = 0, u = 45 m/s and a = -9.8 m/s^2

0 = 45 - 9.8*t

=> t = 45/9.8 = 4.591

The total time taken by the ball is 2*t = 9.18

The time taken by the ball to return from where it was thrown upwards is 9.18 s

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