Answer to Question #153546 in Physics for joy

Question #153546

If the stone enters the water at 41.6 ms-1, and takes 0.6s to travel the 3 meters to the bottom of the pond, what is its average acceleration in the pond water? 




1
Expert's answer
2021-01-04T14:38:39-0500

The distance travelled under a constant acceleration is given as follows:


d=v0tat22d = v_0t -\dfrac{at^2}{2}

where d=3md = 3m is the depth of the pond, t=0.6st = 0.6s is time of movement, v0=41.6m/sv_0 = 41.6m/s is the initial velocity, and aa is the required average acceleration. Expressing aa, find:


a=2v0t2dt2a=241.60.6230.62=122m/s2a = \dfrac{2v_0}{t} - \dfrac{2d}{t^2}\\ a = \dfrac{2\cdot 41.6}{0.6} - \dfrac{2\cdot 3}{0.6^2} = 122m/s^2

Answer. 122 m/s^2.


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