by the energy conservrtion
height it travell is 3.7-1.6=2.1m
K.E initial + P.Einitial = K.Efinal + P.E.Final
0.5mvi2 +mgh1 = 0.5mvf2 + mgh2
m will be cancell from all over the eqaution
0.5*7.52 + 9.8*1.6 = 0.5vf2 + 9.8*3.7
Vf = 3.88458 m/s
yes it reach the top of casstle
part b
speed at top
vf =3.8845 m/s
change in speed=7.5-3.8845=3.6155 m/s
part c
now energy equation
by cancelling m
0.5vtop2 +ghtop = 0.5vground2 + ghlow
0.5*7.52 + 9.8*3.7 = 0.5vground2 + 9.8*1.6
vground =9.8695 m/s
change in speed =9.8695-7.5=2.369 m/s
part d
so it is not same
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