Question
A girl of mass 48.0 kg is rescued from a building fire by leaping into a firefighters net. The window from which she leapt was 12.0 m above the net.She lands in the net so that she is brought to complete stop in 0.45 s.During this interval what is her change in momentum?
Answers
oobleck
mgh = 1/2 mv^2
So, v^2 = 9.81*12*2 = 235.44
her momentum upon impact was p = mv = 48 * 15.34 = 736.51 kg-m/s
So, ∆p = -736.51 kg-m/s
maybe you want the rate of change of p. In that case, just divide by 0.45s
So, v^2 = 9.81*12*2 = 235.44
her momentum upon impact was p = mv = 48 * 15.34 = 736.51 kg-m/s
So, ∆p = -736.51 kg-m/s
maybe you want the rate of change of p. In that case, just divide by 0.45s
Amjad Ali
It was very helpful
Zubair
Good
Anonymous
Solution is not right
Mishi
Good
Anonymous
Wrong solution . your answer is wrong
Aqsa Akram
I think its change in momentum as girl was leaping from height so the initial velocity is zero and final velocity simply can be derived from equation of motion.