A 58.0-kg projectile is fired at anangle of 30.0° above the horizontal with an i

Question

A 58.0-kg projectile is fired at anangle of 30.0° above the horizontal with an initial speed of133 m/s from the top of a cliff164 m above level ground, where theground is taken to be y = 0. (a) What is the initial total mechanical energyof the projectile?
1 J

(b) Suppose the projectile is traveling 85.0 m/s at its maximumheight of y = 360 m. How muchwork has been done on the projectile by air friction?
2 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?
3 m/s (a) What is the initial total mechanical energyof the projectile?
1 J

(b) Suppose the projectile is traveling 85.0 m/s at its maximumheight of y = 360 m. How muchwork has been done on the projectile by air friction?
2 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?
3 m/s

Explanation / Answer

Vy=VoSin-gt, and at the max height vy=0, sot=133Sin30/9.8=6.786 s andy=133Sin30(6.786)-4.9(6.786)2=225.6 plus theinitial height of 164= 389.62. This is the height that theprojectile should have reached, so the height difference is389.62-360=29.62 and energy lost is PE=mgh=(58)(9.8)(29.62)= 16836J. This is just in terms of potential energy, but Vx=115.2m/sshould be the same from the beginning and end of the projectilesmotion, so energy must have been lost in the x direction as well.The speed difference is 115.2-85= 30.2, so the energy lost is KE=1/2(58)(30.2)2=26449 J and the total enegy lost isPEl+KEl˜ 43285 J

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