Question

sitting on the dock of the bay wasting time with my sister. i get bored and push her off the 2 m dock. How fast is she moving when she belly flops into the water? (And more importantly how badly is she going to hurt me when she catches me?)

Answer 1

Answer:

The data that we have is:

Initial height = 2m

Initial vertical speed: As the sister is only pushed, we can assume that we do not have initial speed in the vertical axis.

Then at the beginning, we only have potential energy, that (using the water as the zero in our axis) can be written as:

U = m*g*h

where m is the mass of your sister, g is the gravitational acceleration, and h is the initial height:

h = 2m

g = 9.8m/s^2

Now, when your sister is about to touch the water, all the potential energy has be transformed into kinetic energy:

K = (m/2)*v^2

Then we have, at that moment:

K = U

(m/2)*v^2 = m*g*h

We want to solve this for v, that is the velocity.

v = √(2*g*h) = √(2*9.8m/s^2*2m) = 6.26 m/s

So the speed at which your sister hits the water is 6.26 m/s