Question

A professional diver of mass 60 kg performs a
dive from a platform 10 m above the water
surface. Find the magnitude of the average
impact force experienced by him if the impact
time is 1s on collision with water
surface.Assume that the velocity of the diver
just after entering the water surface is
4 ms-1.(g=10 ms 2​)

Answer 1

Answer:

v=u+at

=4+10(1)

= 14ms

F = m(v-u)/t

= 60(14-4)/1

= 600N

Answer 2

Answer:

The magnitude of the average  impact force experienced is 600 N.

Explanation:

Given that,

Mass of diver = 60 kg

Distance =10 m

Time = 1 sec

Initial velocity = 4 m/s

We need to calculate the final velocity

Using equation of motion

$v=u+gt$

Where, v = final velocity

u = initial velocity

g = acceleration due to gravity

t = time

Put the value into the formula

$v=4+10\times1$

$v = 14\ m/s$

We need to calculate the impact force

Using formula of force

$F=\dfrac{m\Delta v}{t}$

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

Put the value into the formula

$F=\dfrac{60(14-4)}{1}$

$F=600\ N$

Hence, The magnitude of the average  impact force experienced is 600 N.