A 46.0-kg projectile is fired atan angle of 30.0° above the horizontal with an i

Question

A 46.0-kg projectile is fired atan angle of 30.0° above the horizontal with an initial speedof 122 m/s from the top of acliff 135 m above levelground, where the ground is taken to be y = 0. (a) What is the initial total mechanical energy of theprojectile?
J

(b) Suppose the projectile is traveling 85.0 m/s at its maximumheight of y = 294 m. How much work has been done on theprojectile by air friction?
J

(c) What is the speed of the projectile immediately before it hitsthe ground if air friction does one and a half times as much workon the projectile when it is going down as it did when it was goingup?
m/s (a) What is the initial total mechanical energy of theprojectile?
J

(b) Suppose the projectile is traveling 85.0 m/s at its maximumheight of y = 294 m. How much work has been done on theprojectile by air friction?
J

(c) What is the speed of the projectile immediately before it hitsthe ground if air friction does one and a half times as much workon the projectile when it is going down as it did when it was goingup?
m/s

Explanation / Answer

(a)       Initial energyEi = (46kg)(9.8m/s2)(135m) +(1/2)(46kg)(122m/s)2                              = 60858 J + 342332J                               =403190J (b) energy of the object at the maximum height is(46kg)(9.8m/s2)(294m) +(1/2)(46kg)(85m/s)2                                                                          = 298710.2J Then the work done against the air friction willbe  403190J - 298710.2J = 104479.8J (c) In this case Ei = Ef Ef = (1/2)mvf2 Then vf = [(2)(403190J)/(46kg)]            = 132.4m/s Then vf = [(2)(403190J)/(46kg)]            = 132.4m/s

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