Question

A high diver of mass 50.0 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If her downward motion is stopped 2.20 s after her feet first touch the water, what average upward force did the water exert on her?

Answer 1

Let the velocity at first touch the water=v

Answer 2

Now required equation is
v^2=u^2+2gh
v^2=0^2+2*9.8*10
v=?

Answer 3

Negative acceleration within water is =a

Answer 4

Now we wanna use
v=u-at
0=u-a*2.2
Here the value of u will b equal to the value of v calculated b4
a=?

Answer 5

Now use
F=ma=?

Answer 6

Response plz for better understanding!!!"