describe in detail how you could measure the the percentage of elastic energy stored in a catapult which is?

Question

transferred to the k.e of a stone wen the catapult is fired.

2)A skydiver of mas 70kg reaches a speed of 45m/s after falling 150m.by finding the loss of g.p.e and the gain in k.e.determine the work done on the skydiver by the push of the air and hence find the average vertical push of the air on him during the first 150m of his fall.
3)A supermarket troley of total mass 22kg is pushed up a ramp with a force of 14N.The ramp is 5.5m long and rises 0.35m.take g=10N/kg.If the speed of the trolley at the bottom of the ramp was 1.0m/s at the bottom of the ramp,what is its speed at the top.
Practice in physics Pg.40

Answer 1

Find out how far (S) the stone is being hurled of mass m, angle A and how long it took for it to travel (tm)

Kinetic energy Ke
Ke=0.5 mV^2 where
V- speed of the stone
m- mass of the stone

V=sqrt(Vh^2 + Vv^2)
Vh - horizontal velocity
Vv - vertical velocity

and also
Vv=Vh tan(A)
or

V= Vh/cos(A)

A- the angle at which catapult has released the stone

Assume one can measure the time (tm) and distance the stone was hurled

t-total = t + tm
t=sqrt(2h/g) (this is the time it takes to fall from the release point to ground where it fell. )

so
t-total = sqrt(2h/g)+ tm
where
h - height at which the catapult released the stone
g - acceleration due to gravity
and Vh= S/t-total then
V = Vh/cos(A)

V= S/t-total /cos(A)
so kinetic energy is equal to elastic energy and is

E=Ke= 0.5 m [S/t-total /cos(A)]^2

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2) The differnce in energy is the work done by air
Work-air=dE=Pe-Ke
Work-air=mgh - 0.5mv^2
Work-air=m(gh - 0.5v^2)

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3) Pushed up along the ramp or vertical....?
No here on Earth is still g=9.81 m/s^2

Assume parallel to the ramp
Total force is the difference between F= 14N and F-w=mg (h/l)
Ft= F-F-w

S=V0t + 0.5at^2
V=V0+ at

Solve for V