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

intially projectile is at 164m above the ground and its speedis 133 m/s totalenergy=Ke+Pe=1/2mv^2+mgh=58(1/2x133^2+9.8x164)=(6.1x10^5J)amswer (b)at the top of its flight its vertical velocity will be zero,it will have only horizontal velocity.Had there been no friction;horizontal velocity will remain constant = 133cos30=115m/s, but dueto friction its velocity is reduced to 85m/s intial vertical veolocity= 133sin30=66.5 it is given body has reached upto 360m .from ground,as it wasthrown from cliff which is 164 above ground distance travelled byprojectile vertically from the point of throw=360-164=196 we shall calculate the real vertical speed due tofriction using y=196 v^2-u^2=2gs; at the top V=0 so U^2=2x10x196=62.6 due to friction new velocity of projectile is 85i+62.6j magnitude= (85x85+62.5x62.5)=105.5m/s work done by air =1/2m Vi^2- 1/2 mVf^2 =1/2x58(133x133-105.5x105.5)=(1.9x10^5)answer c)work done by air in upward movemnt = 1.9x10e5 ;it is givenair has done 1.5 time more work downward; so total work done by air2.5x1.9x10e5 initial ke of body was=1/2x58x133x133 final ke= 1/2x58xV^2 final ke= intial ke - work done by friction+mgx164 as projectile reaches ground 1/2x58V^2=1/2x58x133x133-2.5x1.9x10e5+58x10x164=5.1x10^5-4.75x10^5+0.9x10^5=1.25x10^5=65m/secanswer V=36m/s

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