A 46.0-kg projectile is fired at an angle of 30.0° above the horizontal with an

Question

A 46.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 124 m/s from the top of a cliff 144 m above level ground, where the ground is taken to be y = 0 (a) What is the initial total mechanical energy of the projectile? (Give your answer to at least three significant figures.) (b) Suppose the projectile is traveling 87.9 m/s at its maximum height of y = 304 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 going down as it did when it was going up? m/s

Explanation / Answer

(a) Initial total mechanical energy = Kinetic Energy (KE) = 1/2 mU^2

= (1/2)*46*124^2 = 353648 J

(b) Suppose the work done by air friction is QE.

Now, total energy at apex is TE = 353648 = KEx + GPE + QE = 1/2 m(U cos(30))^2 + mgy + QE

=> Work done by air friction QE = 353648 - (1/2)*46*(124*cos(30))^2 - 46*9.81*304

= 353648 - 265236 - 137183 = -48771 J

Negative sign shows energy is lost due to air-friction.

(c) Again, we have -

Total energy upon impact is TE = 353648 = KE + QE + 1.5QE = KE + 2.5*QE

=> (1/2) mV^2 = 353648 - 2.5*48771 = 231720.5 J

=> V^2 = (2*231720.5) / 46

=> V = 100.4 m/s

Therefore, the velocity of the projectile immediately before it hits the ground = 100.4 m/s.

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