Question

A high diver of mass 57.8 kg steps off a board 10.0 m above the water and falls vertical to the water, starting from rest. If her downward motion is stopped 1.90 s after her feet first touch the water, what average upward force did the water exert on her? N

Answer 1

the force is described by F=ma

we know her mass and need to find the acceleration

the acceleration will be:

a=(final velocity-initial velocity)/1.90s

the final velocity we are interested in is zero after she is stopped by the water; the initial velocity here is the speed with which she hits the water...

so, we have to find that speed

objects falling a distance h acquire a speed:

v=√[2gh]
here, h=10m

so, v=√[2x9.8x10]=14m/s

therefore, the acceleration is a=-14m/s/1.90s = -7.36 m/s/s

and the force acting on her is

F=57.8 kg x (-7.36 m/s/s)=- 425.40 N
where the minus signs indicate the force is in the direction opposite motion.