Question

What is the force of impact of a 100 kg diver who falls into a swimming pool and hits the surface of the water and comes to a stop in 0.025 seconds? The height is 9 meters

Answer 1

Answer:

F = 53153.36[N]

Explanation:

In order to solve this problem, we must first use the principle of conservation of energy which is transformed from potential energy to kinetic, in this way we can determine the velocity at which the person enters the water.

$E_{pot}=E_{kin}\\m*g*h=\frac{1}{2}*m*v^{2}$

where:

m = mass = 100 [kg]

g = gravity acceleration = 9.81 [m/s²]

h = elevation = 9 [m]

v = velocity [m/s]

Now replacing we can determinate the velocity.

$v^{2}=2*g*h\\v=\sqrt{2*g*h}\\v=\sqrt{2*9.81*9}\\v = 13.28[m/s]$

Then we can calculate the momentum which can be calculated as the product of force by time, this momentum is also equal to the product of mass by velocity.

$P=m*v\\and\\P =F*t\\F*t=m*v$

Now replacing:

F = impact force [N]

t = time = 0.025 [s]

m = 100 [kg]

v = velocity = 13.28 [m/s]

$F*0.025=100*13.28\\F=53153.36[N]$