A 60.0-kg projectile is fired at an angle of 30.0° above the horizontal with an

Question

A 60.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 1.32 102 m/s from the top of a cliff 120 m above level ground, where the ground is taken to be y = 0.
(a) What is the initial total mechanical energy of the projectile?




(b) Suppose the projectile is traveling 93.5 m/s at its maximum height of y = 301 m. How much work has been done on the projectile by air friction?
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(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) So initially the particle has energy = 1/2 mv^2+ mgh= 0.5 *60*132^2+60*9.81*120=593352J b) the energy at maximum height = 1/2 mv^2+ mgh= 0.5 *60*93.5^2+60*9.81*301=439436.1 J the work done by air resistance is the change in energy= 439436.1-593352=-153915.9 J c) So air resistance will do another 1.5*-153915.9=-230873.85 J of work so total energy=439436.1-230873.85=208562.25 so 1/2 mv^2 + mg (0)=208562.25 so v= sqrt(2*208562.25/60)=83.38 m/s

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