m•g•h =m•v²/2,
v² =2•g•h,
a=v²/2•s=2•g•h/2•s= g•h/s=9.8•10/5=2•g=19.6 m/s².
F=m•a= m•2g= 2(m•g)=2•62•19.6 = 1215.2 N.
an athlete drops from rest from a platform 10m above the surface of a 5m deep pool. Assume that the athlete enters the water vertically and moves through the water with the constant acceleration, what is the minimum average force with water must exert on a 62 kg diver to prevent her from hitting the bottom of the pool? Express answer in newtons and also in multiples of the divers weight
2 answers
You have to add the fores in the air and the water
Air: a=9.8 m/s^2 due to gravity
Force=ma, F=62(9.8)=607.6
Water:
First find the velocity in which the diver enters the water,
V^2=V0^2 + 2ay
V^2= 0 + 2(9.8)(10)
V= sqr(20*9.8)= 14m/s
Now use velocity to solve for acceleration through the water
V^2=V0^2 + 2ay
0=14^2 + 2a(5)
-14^2= 10a
a=19.6 m/s^2
Now find force through water
F=ma
F=62(19.6)=1215.2
Now add force through water and force through air
1215.2+607.6=1822.8 N
Air: a=9.8 m/s^2 due to gravity
Force=ma, F=62(9.8)=607.6
Water:
First find the velocity in which the diver enters the water,
V^2=V0^2 + 2ay
V^2= 0 + 2(9.8)(10)
V= sqr(20*9.8)= 14m/s
Now use velocity to solve for acceleration through the water
V^2=V0^2 + 2ay
0=14^2 + 2a(5)
-14^2= 10a
a=19.6 m/s^2
Now find force through water
F=ma
F=62(19.6)=1215.2
Now add force through water and force through air
1215.2+607.6=1822.8 N
0 answers