Asked by Anonymous
A 56 kg swimmer jumps off a 15 m tower. Find the swimmer's velocity when hitting the water. The swimmer comes to a stop 2 m below the surface. Find the net force exerted by the water.
Answers
Answered by
Natalie
In the air:
Vi=0, d=15m, a=9.8m/s^2, vf=?
vf^2=vi^2+2ad
vf^2=0^2+2*9.8*15
vf=17.15m/s --> speed when hitting the water
Now for the water part:
vi=17.15m/s, vf=0,d=2,a=?
vf^2=vi^2+2ad
0^2=17.15^2+2*a*2
a=-73.5 m/s^2
F=ma
F=56*73.5=4116 N
Vi=0, d=15m, a=9.8m/s^2, vf=?
vf^2=vi^2+2ad
vf^2=0^2+2*9.8*15
vf=17.15m/s --> speed when hitting the water
Now for the water part:
vi=17.15m/s, vf=0,d=2,a=?
vf^2=vi^2+2ad
0^2=17.15^2+2*a*2
a=-73.5 m/s^2
F=ma
F=56*73.5=4116 N
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