Question

A 100 kg man jumps into a swimming pool from a height of 5.0 meters. It takes 0.40 seconds for the water to reduce his velocity to zero. what average force did the water exert on the man?

Answer 1

Force exerted by the water = 3500 N.

Solution enclosed.

Answer 2

Answer:

m=100kg

h=5m

t=0.4 sec

v= under root (2gh)

$\sqrt{2 \times 10 \times 5}$

10 m/sec

net force on man in upward direction

m=v-u/t

m=-u/t

=10 ×-100/0.4

-2500N

downward weight =100 ×10

=1000N

Total force exerted by water =3500N

ANSWER.