A parachutist and parachute have a combined mass of 93 kg. They are travelling vertically downwards at a constant speed of 0.30 m.s−1. What is the magnitude of the force of air resistance on them?
Answers
Explanation:
As per the question,
Velocity of the ball =100kph=100\times\dfrac{1000}{60\times 60}=\dfrac{1000}{36} m/sec100×60×601000=361000m/sec
Mass of the ball=0.16Kg
duration of collision =1ms=1\times10^{-3}1×10−3 sec
Force = Rate of change of momentum=\dfrac{mv-(-mv)}{t}=\dfrac{2mv}{t}tmv−(−mv)=t2mv
Now, substituting the values in the above equation,
F=\dfrac{2\times0.16\times1000}{36\times1\times 10^{-3}}=8.89\times 10^{3}NF=36×1×10−32×0.16×1000=8.89×103N
b)
Mass of the Parachutist and parachute(m)=93kg
downwards Speed(v) =0.3m/sec=0.3m/sec
Let the resistive force =F_sFs
and the gravitational acceleration =g
As per the question, they are moving with the constant speed,
So, net force must be zero,
mg-F_s=0mg−Fs=0
F_s=mg=93\times 9.8 N=911.4NFs=mg=93×9.8N=911.4N
If I used F=ma then that would imply there's acceleration, but there's still velocity and mass which affect the air resistance. I'm not sure how to take the velocity into account and which equation to use. If I'm understanding how terminal velocity works, then air resistance would equal weight (zero sum on the free body diagram) but this looks like the parachute opened already.