trampoline girl..?
A girl of mass m1 = 60 kilograms springs from a trampoline with an initial upward velocity of vi = 8.0meters per second. At height h=2.0 meters above the trampoline, the girl grabs a box of mass m2 = 15 kilograms.
For this problem, use g = 9.8 meters per second per second for the magnitude of the acceleration due to gravity.
a) What is the speed v-before of the girl immediately before she grabs the box?
b) What is the speed v-after of the girl immediately before she grabs the box?
c) What is the maximum height h-max that the girl (with box) reaches? Measure h-max with respect to the top of the trampoline.
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Kia ora
It seems reasonable to assume that the box is stationary at the "time of grabbing".
a) the speed of the girl 2.0m above the trampoline is most easily found using a kinematic relationship.
v²=u²+2as
v=√(8.0²+2*-9.8*2.0)=4.98m/s
b) I guess you mean "just after she grabs the box"?
Using conservation of momentum, the momentum of the girl+box after she grabs it will be equal to her momentum just before she grabs it.
M=mass of girl
m=mass of box
u=initial velocity of girl
v=final velocity of girl+box
Mu=(M+m)v
v=Mu/(M+m)
v=60*4.98/(60+15)
v=3.98m/s
c) Using v²=u²+2as again:
v=0
u=3.98
a=-9.8
s=?
s=(v²-u²)/2a
s=(0-15.8)/-19.6
s=0.81m
Add the 2.0m from the top of the tramp gives a total height of 2.8m.
You cannot use conservation of energy as the collision ("grabbing") is inelastic.
Trampoline Box
If you recognize what "the person exhibit" is.......good it consistently ended with women leaping on trampolines. But yeah.....with no trouble leaping on a trampoline is not haram. Doing it out for the sector to observe could be.
You need the velocity of the box immediately before she grabs it to answer the question.