A swimmer dives from a height of 25 m into water. If the swimmer jumps forward with a speed 8 m/s, how far from the diving board does he land into the water.



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Posted on (Answer #1)

The swimmer dives from a point 25 meters above water level. The velocity of the swimmer is 8 m/s in the horizontal direction. The velocity in the vertical direction is 0 m/s. As the swimmer falls, there is a downward acceleration of 9.8 m/s^2.

To determine the time taken by the swimmer to fall in the water use the formula `S = u*t + (1/2)*a*t^2` where S is the distance traveled by a body in time t if the initial velocity is u and the acceleration is a.

`25 = 0 + (1/2)*9.8*t^2`

=> `t = sqrt(25/4.9)`

=> `t = (5/7)*sqrt 10 ~~ 2.25` s

In this time, the horizontal displacement of the swimmer is 18.07 m.

The swimmer lands in the water 18.07 m away from the point where he dives from.

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