A 56.0-kg projectile is fired at an angle of 30.0° above the horizontal with an

Question

A 56.0-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 118 m above level ground, where the ground is taken to be y = 0.

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


(b) Suppose the projectile is traveling 86.5 m/s at its maximum height of y = 273 m. How much work has been done on the projectile by air friction 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 in m/s?

Explanation / Answer

a) total mechanical energy = mgh + 1/2 mv^2 = 356 kJ b) initial horizontal velocity = ucos30 = 102 cos 30 = 88.3 m/sec let f be the air frictional force, u1 be the initial upward velocity a be the acceleration in upward direction a= (mg+f)/m v^2 - u1^2 = 2aS 0- (102sin30)^2 = -2 (mg+f)/m * S from this we get f= -78.9 N

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