A 44.0 kg projectile is fired at an angle of 30.0 degrees above the horizontal w

Question

A 44.0 kg projectile is fired at an angle of 30.0 degrees above the horizontal with an initial speed of 1.4 x 102 m/s from the top of a cliff 158m above the level ground, where the ground is taken to be y=0.

a) What is the initial total mechanical energy of the projectile? (Answer in J)

b) Suppose the projectile is traveling 99.2 m/s at its maximum height of y=362m. How much work has been done on the projectile by air friction? (Answer in 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? (Answer is m/s)

Explanation / Answer

Given that mass of the projectile is m = 44.0kg initial speed is v_i = 140 m/s^2 Height of the cliff is   h = 158 m the initial total mechanical energy of the projectile is E = K.E + P.E       = 0 + mg h       = 44.0 kg * 9.8 m/s^2 * 158 m       = 6.81 *10^4 J b ) Suppose the projectile is travelling at a speed is v_i = 99.2 m/s Work done is K.E_f - K.E_i        1/2 m v_i^2 -1 /2 m v_f^2    = 1/ 2 *44.0 kg * ( 99.2 m/s)^2 - 1/2 44.0 kg ( 140 m/s)^2 = - 2.147*10^5 J c ) Work done is W = 3/2 W       mg h - 3/2 W = 1/2 m v^2         44.0kg * 9.8 m/s^2 * 158 m - 3/ 2 * (- 2.147*10^5 J ) = 1/2 * 44.0 kg * v^2        v = 213.5 m/s

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