Posted by Einstein on .
A 65 kg dancer leaps .32 m high.
a) With what momentum does the dancer reach the ground?
b) What impulse is needed to make a stop?
c) As the dancer lands, the knees bend, lengthening the stopping time to .050 s. Find the average force exerted on the body.
d) Compare the stopping force to the performer's weight.
Thoughts: Momentum equals mass multiplied by the change in velocity. Impulse is the force multiplied by time.
a) Use energy conservation considerations to compute the takeoff velocity.
(1/2) M V^2 = M g H
V = sqrt (2gH)
The landing velocity will be the same. Multiply that V by M for the momentum
(b) Same number as (a)
(c) Force = Impluse/Time
(d) Compare (c) answer to Mg
A) Momentum = (2.465)(65) = 160.222 kgm/s
B) 160.222 kgm/s
C) Force = 160.222/.050 = 3204.441 N
D) 3204.441/[(65)(9.8)] = 5.031