mgh=mv₀²/2
v₀=sqrt(2gh)=sqrt(2•9.8•10.4)=14.28 m/s
v= v₀-at, v=0 =>
a= v₀/t = 14.28/3.58 =4 m/s²
F=ma=68.2•4=272.8 N
A high diver of mass 68.2 kg jumps off a board 10.4 m above the water.
The acceleration of gravity is 9.8 m/s2 .
If his downward motion is stopped 3.58 s after he enters the water, what average upward force did the water exert on him?
I already calculated that his velocity before hitting the water is 14.277 m/s.
1 answer
1 answer