A high-diver of mass 59.8 kg jumps off a board 10.0 m above the water.If,1.00 s after entering the water his downward motion is stopped,what average upward force did the water exert?

Question

Quizplus · Accepted Answer

First, calculate the diver's velocity just before entering the water using the equation for free fall: $v^2 = u^2 + 2gh$, where $u = 0$ (initial velocity), $g = 9.8 \, \text{m/s}^2$ (acceleration due to gravity), and $h = 10.0 \, \text{m}$. This gives $v = \sqrt{2 \times 9.8 \times 10.0} = \sqrt{196} = 14 \, \text{m/s}$.The change in momentum ($\Delta p$) of the diver when he stops is given by $\Delta p = mv - mu$, where $m = 59.8 \, \text{kg}$, $v = 0 \, \text{m/s}$ (final velocity), and $u = 14 \, \text{m/s}$ (initial velocity). So, $\Delta p = 59.8 \times 0 - 59.8 \times 14 = -838.8 \, \text{kg m/s}$.The average force ($F$) exerted by the water can be found using the impulse-momentum theorem, $F \Delta t = \Delta p$, where $\Delta t = 1.00 \, \text{s}$. Rearranging for $F$, we get $F = \frac{\Delta p}{\Delta t} = \frac{-838.8}{1.00} = -838.8 \, \text{N}$.The negative sign indicates the direction of the force is opposite to the direction of motion (upward in this case), but since the question asks for the magnitude of the force, the answer is $838.8 \, \text{N}$, which is not listed among the options, making E (None of the other answers is correct) the correct choice.

A High-Diver of Mass 59