A boy jumps vertically upwards fromthe top of a platform and then drops onto a trampoline which is 3m below his start point.
If the trampoline has an effective spring constant 5.2x10^4N/m
and is depressed 0.35m, calculate the boy's initia; upward
velocity.
let initial velocity be v.
then initial energy=0.5*m*v^2(assume potential energy at this height is 0.then if we go up,potnetial energy will increase and if we go down,energy will decrease.)
so as spring is depressed, total depth=3.35 m
so at this depth,total energy=-m*g*3.35+0.5*5.2*10^4*0.35^2
so equating this energy with initial kinetic energy,
we get
v=79.4 m/s
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