Question

A high diver of mass 63.0 kg Jumps off a board 10.0 m above the water. If her downward motion is stopped 1.80 s after she enters the water, what average upward force did the water exert on her?

Answer 1

we want to find the speed of the diver on hitting the water, so we use

acceleration = change in velocity / time

then , we use F = m*a to find force

if his initial vertical velocity was zero, then his speed hitting the water is given by:

vf^2 = v0^2 + 2*a*d

vf = final velocity

v0 = initial velocity = 0

a = 9.8 m/s^2

d=10 m

vf = sqrt [2*9.8 *10] =14m/s

now, we apply this equaiton again to get his acceleration caused by the water:

acceleration = change in velocity / time

a = (0 - 14) /1.80 = 7.77 m/s^2

therefore, F = m*a = 63 * (-7.77) =-490 N

the minus signs indicate the force acts opposite to the direction of motion

m*g = 63*9.81

average upward force = 490 + 63*9.81

average upward force = 1107.4 N