Question

A 68.0 kg diver falls from rest into a swimming pool from a height of 4.90 m. It takes 1.46 s for the diver to stop after entering the water. Find the magnitude of the average force exerted on the diver during that time.

Answer 1

Answer:

456.4 N

Explanation:

From conservation of energy, the potential energy is converted to kinetic energy hence

mgh=½mv²

Making v the subject then

$v=\sqrt {2gh}$

Where g is acceleration due to gravity and h is height whike v is final velocity. Substituting 4.9m for h and 9.81 m/s² for g then

$v=\sqrt {2\times 9.81\times 4.9}=9.80499872514015m/s\approx 9.8 m/s$

Change in momentum equals to the impulse.

Impulse, I= Ft

Change in momentum, ∆p= m(v-u)

Ft=m(v-u) making F the subject of formula then

$F=\frac {m(v-u)}{t}$

Where F is force in Newton, t is time in seconds, m is mass of diver, v and u are the final and initial velocities respectively.

Substituting 68 kg for m, 9.8 m/s for v, 0 m/s for u since it is initially at rest and 1.46 s for t

$F=\frac {68(9.8-0)}{1.46}=456.43835616438N\approx 456.4N$