Question

A 80 kg man drops from a diving board above the surface of the water of a swimming pool. When he reaches the water, it applies an upward force of 468N over a time interval of 0.68s bringing him to a complete stop. What was the velocity of the man when he made contact with the water? (As the man is headed down assign the velocity as a negative number)

Answer 1

Given data

*The given mass of the man is m = 80 kg

*The given upward force is F = 468 N

*The given time is t = 0.68 s

The formula for the velocity of the man is calculated by the impulse-momentum theorem as

`\begin{gathered} F* t=m\Delta v \\ \Delta v=(F* t)/(m) \end{gathered}`

Substitute the known values in the above expression as

`\begin{gathered} \Delta v=((468)(0.68))/((80)) \\ =3.97\text{ m/s} \end{gathered}`

Hence, the velocity of the man when he made contact with the water is 3.97 m/s