An 70 kg man drops from rest on a diving
board −2.9 m above the surface of the water
and comes to rest 0.58 s after reaching the
water.
The acceleration due to gravity is
9.81 m/s
2
.
What force does the water exert on the
man?
Answer in units of N
3 answers
I think it would just be Fg x m, so 70kg times 9.81 m/s^2, and then multiply all of that by 2.9m, and the negative is just saying what direction the man is falling, meaning downwards
mgh=mv²/2
v=sqrt{2gh} =sqrt{2•9.8•2.9} =7.5 m/s
v(fin)=v-at
v(fin)=0
a=v/t
x=vt -at²/2=vt/2
mv²/2 =W(fr) = F(fr) •x =>
F(fr)= mv²/2x = 2mv²/2 vt=mv/t =
=70•7.5/0.58 = 905.2 N
v=sqrt{2gh} =sqrt{2•9.8•2.9} =7.5 m/s
v(fin)=v-at
v(fin)=0
a=v/t
x=vt -at²/2=vt/2
mv²/2 =W(fr) = F(fr) •x =>
F(fr)= mv²/2x = 2mv²/2 vt=mv/t =
=70•7.5/0.58 = 905.2 N
The above equation is wrong