Question

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?

Answer 1

`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`.