A 65.0-kg circus performer is fired from a cannon that is elevated at an angle of 55.8 ¡ã above the horizontal. The cannon uses strong elastic bands to propel the performer, much in the same way that a slingshot fires a stone. Setting up for this stunt involves stretching the bands by 2.12 m from their unstrained length. At the point where the performer flies free of the bands, his height above the floor is the same as that of the net into which he is shot. He takes 2.72 s to travel the horizontal distance of 24.7 m between this point and the net. Ignore friction and air resistance and determine the effective spring constant of the firing mechanism.

Question

Damon · Accepted Answer

LOL, the way MathMate did it is a lot easier :)

MathMate · Answer

See earlier post: http://www.jiskha.com/display.cgi?id=1400921781

Damon · Answer

The total energy at the top of the trajectory is the same as was stored in the spring (1/2)kx^2

Total energy at top where v = 0 = (1/2) u^2 + m g h

we can find u, horizontal velocity easily
u = 24.7 m/2.72 s = 9.08 m/s

but Vi, initial speed up is easy too
tan 55.8 = Vi/u
Vi = 9.08 * tan 55.8 = 13.4 m/s

so we need to find where v = 0 to get the height
v = Vi - 9.81 t
0 = 13.4 - 9.81 t
t = 1.36 seconds to top
(which is 2.72 / 2 of course. It goes up half the time :)

so how high in 1.36 seconds?
h = Vi t - 4.p t^2
= 13.4(1.36) - 4.9 (1.36^2)
= 18.22 - 9.06 = 9.1 meters high at top
potential energy at top = m g h
= 65 * 9.81 * 9.1 = 5802 Joules
kinetic energy at top = (1/2)65(9.08^2)
= 2680
total energy = 8480 Joules
(1/2) k 2.12^2 = 8480
k = 3774 N/m