A 52.0-kg projectile is fired at an angle of 30.0° above the horizontal with an

Question

A 52.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 1.22 102 m/s from the top of a cliff 148 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 86.5 m/s at its maximum height of y = 303 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?

Explanation / Answer

a)initial mechanical energy = 52*9.8*148+(1/2)*52*122^2 = 75420.8 + 386984 = 462404.8 J b)work done by air friction = 462404.8 - (0.5*52*86.5^2 + 52*9.8*303) = 462404.8 - 348947.3 = 113457.5 J c)work done by friction=3/2*113457.5 = 170186.25 final energy = 348947.3 - 170186.25 = 178761.05 J 1/2 * 52 * v^2 = 178761.05 J => v = 82.92m/s

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