A 58.0 -kg projectile is fired at an angle of 30.0 A 58.0-kg projectile is fired

Question

A 58.0-kg projectile is fired at an angle of 30.0

A 58.0-kg projectile is fired at an angle of 30.0 degree above the horizontal with an initial speed of 1.38 times 102 m/s from the top of a cliff 156 m above level ground, where the ground is taken to be y = 0. What is the initial total mechanical energy of the projectile? Suppose the projectile is traveling 97.8 m/s at its maximum height of y = 354 m. How much work has been done on the projectile by air friction? 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 going down as it did when it was going up?

Explanation / Answer

a) total energy = potential + kinetic = mgh +0.5 mv^2 = 640.94 x10 ^3 J

b) work done by friction is energy at highest point - intial = (640.94 -478.9 ) x 10 ^ 3J = 162.34 x 10 ^ 3

c) since it lossed 1.5 times while coming back ,total energy lost = 2.5 x 162.34 x 10 ^ 3 = 405.86 x10 ^3

finally there is only kinetic energy ,0.5x m x v ^2 = 640 .94 -405.6 kJ

   V = 90.08 m / sec

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