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

Question

A 60.0-kg projectile is fired at anangle of 30.0° above the horizontal with an initial speed of120 m/s from the top of a cliff145 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 = 290 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 = 290 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

(a) What is the initial total mechanical energyof the projectile?

Et = Pe + Ke = mgh + 0.5mV2 Checked!
(b) Suppose the projectile is traveling 85.0 m/s at its maximumheight of y = 290 m. How muchwork has been done on the projectile by air friction?
Wair +Wgravity + Keat h-max =Keinitial Wair = Keinitial -Wgravity - Keat h-max (at h-max wehave only the horizontal component of V; where Vh= Vsos(30.0)) Wair = 0.5mVinitial2 - mghmax   -0.5mVh2 or Wair = 0.5m(Vinitial2 - 2ghmax   -Vh2 ) where hmax = 290 m Checked! (c) What is the speed of the projectileimmediately before it hits the ground if air friction does one anda half times as much work on the projectile when it is going downas it did when it was going up? Keinitial + Pe=Wair-up + Wair-down +Kefinal   since   Wair-down =1.5Wair-up  we have Keinitial + Pe =Wair-up +1.5Wair-up  +Kefinal   Keinitial + Pe = 2.5Wair-up  +Kefinal   Kefinal  =Keinitial - 2.5Wair-up  + Pe 0.5mV2 = 0.5mVinitial2 -  2.5Wair-up   + Pe   [Pe= Pe initial] since Wair = 0.5m(Vinitial2 - 2ghmax   -Vh2 ) and Pe = mgh wehave . 0.5mV2 = 0.5mVinitial2 -  2.5 (0.5m(Vinitial2 - 2ghmax   -Vh2 ) ) + mgh 0.5mV2 = 0.5mVinitial2 -    5/4 m( Vinitial2 -2ghmax   - Vh2 ) +mgh 0.5V2 = 0.5Vinitial2 -    5/4 (Vinitial2 - 2ghmax   -Vh2 ) + gh V2 = Vinitial2 -    5/2 (Vinitial2 - 2ghmax   -Vh2 ) + 2gh V2 = -3/2Vinitial2    +5ghmax   + 5/2Vh2 + 2gh V2=  5ghmax   + 5/2Vh2 + 2gh  - 3/2Vinitial2 V=  [5ghmax   +5/2Vh2 + 2gh  -3/2Vinitial2] hmax= 290 m h= 145 m V= 120m/s Checked!

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