a 60 kg dancer leaps 0.32 m. high.?
a) with what momentum does the dancer reach the ground?
b) what impulse is needed to stop the dancer?
c) as the dancer lands, his knees bend, lengthening the stopping time to 0.050 s. find the average force exerted on the dancer’s body.
b) what impulse is needed to stop the dancer?
c) as the dancer lands, his knees bend, lengthening the stopping time to 0.050 s. find the average force exerted on the dancer’s body.
Comments(1)
a) Kinetic energy at ground = Potential energy at height
(1/2) mv^2 = mgh or v=sqrt(2 g h) = sqrt(2 x 9.813 x 0.32) = 2.5 m/s
b) Total momentum of dancer = 60 x 2.5 = 150 kg m/s
c) Average force = momentum change / time = 150/0.05 = 3000 N