A 65-kg swimmer jumps off a 10.0-m tower. What is the swimmer's velocity on hitting the water?

The swimmer comes to a stop 2.0 m below the surface. What is the net force exerted by the water?

1 Answer
Jan 17, 2018

v = 14m/s and F_(Net) = 2548N upwards

Explanation:

First, remember the equation v_f^2 - v_i^2 = 2ad.
We want to find v_f and we know that the swimmer is stationary so v_i would be 0m/s.

So our equation would be v_f^2 = 2ad and since we know a which is 9.8m/s^2 or g and that d = 10m it is a matter of plugging the numbers in:
v_f^2 = 2(9.8m/s^2)(10m).
So v_f = sqrt(196m^2/s^2) = 14m/s.

Now that we know the velocity, we can find the answer to the second part.

We can still use the same equation again, except that our v_f is equal to 0 and that v_i = 14m/s. We want to find a, which is the unknown and we know d. So plugging in what we know, we will get:
-196m^2/s^2 = 2a(2m).

Isolating a, we would find that a = -49m/s^2 or a = 49m/s^2 if you consider upwards to be positive.

Then finding F_(water), we would get that it is equal to 3185N. Then considering that there is F_g pushing the swimmer down, we would get F_g = 637N.

So the net force pushing the swimmer up will be F_(water) - F_g = 2548N.