Answer:
a) 39.6 m/s b) 4123 N
Explanation:
a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).
Fnet=ma
ma=m(v^2/R) (centripetal acceleration)
mg=m(v^2/R)
m cancels out (this is why pilot feels weightless) so,
g=(v^2/R)
9.8 m/s^2 = v^2/160 m
v^2=1568 m^2/s^2
v=39.6 m/s
b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.
Convert 300 km/hr to m/s
300 km/hr=83.3 m/s
Convert pilot's weight into mass:
760 N = 77.55 kg
Fnet=ma
n-mg=m(v^2/R)
n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)
n=3363.2 N+760 N=4123 N
Hmm doesnt soujd familiar
<span>K.E = 0.5 * m * v^2 ( m = mass(Kg), V = Velocity(m/s)
= 0.5 * 8 * 5^2
= 4 * 25
= 100 J </span>
Explanation :
The forces acting on hot- air balloon are:
Weight, (W)
Force due to air resistance, (F)
Upthrust force, (U)
Its weight W is acting in downward direction. The upthrust force U acts in upward direction. When the balloon is moving upward, the air resistance is in downward and vice versa.
In this case, the hot-air balloon descends vertically at constant speed.
so, 
and 
so,
....................(1)
when it is ascending let the weight that it is releasing is R, so
..........(2)
solving equation (1) and (2)

2F is the weight of material that must be released from the balloon so that it ascends vertically at the same constant speed.