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#.