A 44.0-kg projectile is fired at an angle of 30.0° above the horizontal with an

Question

A 44.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 1.34 102 m/s from the top of a cliff 152 m above level ground, where the ground is taken to be y = 0. (a) What is the initial total mechanical energy of the projectile? J (b) Suppose the projectile is traveling 95.0 m/s at its maximum height of y = 339 m. How much work has been done on the projectile by air friction? J (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 going down as it did when it was going up? m/s

Explanation / Answer

a.) Initial total energy = Kinetic energy + Potential Energy = 0.5 mv2  + mgh = 0.5 x 44x 1.341022 + 44x9.81 x 152

= 65648.84336 J

b.) Gain in total energy is the work done by air resistance.

Work done by air = total energy at highest point - Initial Total energy

= (Kinetic energy at hghest point + Potential energy at highest point) - 65648.84336

=   ( 0.5 x 44 x 952 + 44 x 9.81x 339 ) - 65648.84336 = 344875.96 - 65648.84336 = 279227.1166 J

c.) If air does 1.5 times the work, then increase in total energy = 1.5 x 279227.1166 = 418840.675 J

So, total energy of the body at the lowest point when it is just about to hit the ground

= Energy at the highest point + 418840.675 = 344875.96 + 418840.675 = 763716.635 J

Since the potential energy at the ground is zero, All of it is KE.

So, 0.5 mV2 = 763716.635

0.5 x 44 x V2 = 763716.635

V = 186.3179876 m/s

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