A 50.0-kg projectile is fired at an angle of 30.00 above the horizontal with an
ID: 1359543 • Letter: A
Question
A 50.0-kg projectile is fired at an angle of 30.00 above the horizontal with an initial speed of 1.20 x 102 m/s from the top of a cliff 142 m above level ground.
What is the initial total mechanical energy of the projectile?
Suppose the projectile is traveling 85.0 m/s at its maximum height of 427 m. How
much work has been done on the projectile by air resistance?
What is the speed of the projectile immediately before it hits the ground if air
resistance does 1.5 times as much work on the projectile when it’s going down as it did when it was going up?
Explanation / Answer
given,
mass = 50 kg
speed = 120 m/s
height = 142 m
total mechanical energy = mgh + 0.5 * mv^2
total mechanical energy = 50 * 9.8 * 142 + 0.5 * 50 * 120^2
total mechanical energy = 429580 J
if projectile is traveling 85.0 m/s at its maximum height of 427 m its total mechanical energy will be
total mechanical energy = 50 * 9.8 * 427 + 0.5 * 50 * 85^2
total mechanical energy = 389855 J
work done by air resistance = 429580 - 389855
work done by air resistance = 39725 J
when it reaches ground its potential energy will be 0 so
by conservation of energy
initial energy = final energy
429580 = 0.5 * m * v^2 + 39725 + 39725 * 1.5
429580 = 0.5 * 50 * v^2 + 39725 + 39725 * 1.5
v = 114.938 m/s
speed of the projectile immediately before it hits the ground = 114.938 m/s
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