12. -13 points SerCP11 5.P029. My Notes Ask Your Teacher A 56.0-kg projectile is
ID: 1659145 • Letter: 1
Question
12. -13 points SerCP11 5.P029. My Notes Ask Your Teacher A 56.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 130 m/s from the top of a cliff 114 m above level ground, where the ground is taken to be y (a) What is the initial total mechanical energy of the projectile? (Give your answer to at least three significant figures.) (b) Suppose the projectile is traveling 92.1 m/s at its maximum height of y=290 m. How much work has been done on the projectile by air friction? (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? m/sExplanation / Answer
(a)
Total Mechanical Energy :
KE = (1/2)*56*130^2 = 473200 J
PE = 56*9.8*114 = 62563.2 J
Total = 535763.2 J
(b)
KE = (1/2)*56*92.1^2 = 237507.48 J
PE = 56*9.8*290 = 159152 J
Total = 396659.48 J
Horizontal Velocity = 130 cos(30)
Kinetic Energy = (1/2)*56*(130cos(30))^2 = 354900
Work = 354900 - 237507.48 = 117392.52 J
(c)
Work = 1.5 * 117392.52 = 176088.78 J
At ground the PE will be 0 and it will all be converted to KE.
KE = 396659.48 – 176088.78 = 220570.7 J
(1/2)*56*v^2 = 220570.7
v = 88.7554 m/s
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