A standing 80 kg man steps off a 2.9 m high diving platform and begins to fall from rest. The man comes to rest 2.0 seconds after reaching the water. What average force did the water exert on him?

Respuesta :

Answer:

[tex]Force=1085.56N[/tex]

Explanation:

Given data

Mass m=80 kg

Distance s=2.9 m

Time t=2.0 s

To find

Average Force

Solution

First we need to find the velocity

So

[tex]V^{2}=2as\\V^{2}=2(9.8m/s^{2} ) (2.9m)\\V^{2}=56.84\\V=\sqrt{56.84}\\ V=7.539m/s[/tex]

At 2.0 seconds the force must first overcome gravity

So

Force due to gravity=mass×gravity

                                 =(80kg)×(9.8m/s²)

                                 =784N

To Slow him F×t=mV

[tex]F*t=mV\\F=(mV)/t\\F=(80kg*7.539m/s)/2.0s\\F=301.56N[/tex]

The Net Force is given as:

[tex]Force=301.56N+784N\\ Force=1085.56N[/tex]