Question

A diver of 50 kg jumps from a platform 20 m, high into a pool. If the diver comes
to rest in 0.8 seconds after hitting the surface of the water, then what is the
force that the water exerts on the diver? (g = 10 m/s2)
625 N
875 N
1175 N
1250 N​

Answer 1

Answer:

1175n.........................

Answer 2

Answer:

1250 N

Explanation:

It takes the diver 2 seconds to fall to the surface of the water because acceleration due to gravity is 10m/s^2 and the diver is passing 20m of distance, so the acceleration at the surface of the water is 20m/s.

Once it hits the surface of the water, the diver deccelerates to 0m/s in 0.8 seconds.
So to find the acceleration, we do △v/t, or (vf - vi)/t.
(0-20)/0.8 is -25m/s^2.

To find the force of the water which caused the diver to slow down, we use Force = mass * acceleration.

F = 50kg * 25m/s^2 = 1250 N.