If we define the initial velocity as having magnitude uu, then the vertical component of this is u*sin30u⋅sin30 (the horizontal component would be u*cos30u⋅cos30 but we are not interested in this as it has no effect on potential energy)
A word of caution: there is no definition or reference in the question as to what 30 degrees is measured from; it could be from the horizontal or the vertical. | have assumed from the horizontal. Please let me know if this is not the case and I will re-work.
The body will travel a maximum distance ss upwards. At this point its vertical velocity will be nil. We can used the following to determine ss:
v^2=u^2+2asv2=u2+2as
We know vv, final velocity is nil, and uu, initial velocity is usin30usin30 and aa will be acceleration due to gravity, gg (which will be negative relative to the direction of the initial velocity). Hence:
0=u^2sin^2 30-2gs0=u2sin230−2gs
2gs=u^2sin^2 302gs=u2sin230
s=(u^2sin^2 30)/(2g)s=u2sin2302g
So final potential energy will be:
PE=mgh=mgsPE=mgh=mgs
PE=(mg)*(u^2sin^2 30)/(2g)PE=(mg)⋅u2sin2302g
PE=(m*u^2sin^2 30)/(2)PE=m⋅u2sin2302
Initial kinetic energy is
KE=1/2m*u^2KE=12m⋅u2
So percentage conversion is:
"PE"/"KE"=(m*u^2sin^2 30)/(2)/(1/2m*u^2)PEKE=m⋅u2sin2302/(12m⋅u2)
"PE"/"KE"=sin^2 30=0.25PEKE=sin230=0.25
So the answer is 25%