A50 kg projectile is fired at an angle of 30 degrees above the horizontal with an initial speed of 1.2 x 10^2 m/s from the top of a cliff 142 m above level ground, where the ground is taken to be y=0. a) what is the initial total mechanical energy of the projectile? b) suppose the projectile is traveling 85 m/s at its max height of y=427 m. how much work has been done on the projectile by air friction? c) what is the speed of the projectile immediately before it hits the ground if air friction does one and a half times as much work on the projectile when it is go down as when go up?

A) The total mechanical energy is:
M.E = Kinetic energy + Potential energy
M.E = 1/2 mv² + mgh
= 0.5 x 50 x 120² + 50 x 9.81 x 142
= 430,000 J

b) Work done by friction will be equal to the mechanical energy lost
M.E at max height: 1/2 x 50 x 85² + 50 x 9.81 x 427
= 390,000 J
Change in ME = work done by air =
430000 - 390000
= 40,000 J

c) All of the potential energy will be converted to kinetic energy and added to its existing kinetic energy at maximum height. Also, the work done by air will be subtracted from the M.E because it will be lost.
K.E(final) = M.E(max height) - 1/2 (Work done by air while going up)
1/2 50v² = 390,000 - 20,000
v = 121.7 meters per second

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