A 52.0-kg projectile is fired at an angle of 30.0° above the horizontal with an

Question

A 52.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 1.24 102 m/s from the top of a cliff 148 m above level ground, where the ground is taken to be y = 0.
(a) What is the initial total mechanical energy of the projectile?
(b) Suppose the projectile is traveling 87.9 m/s at its maximum height of y = 308 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?

Explanation / Answer

intially projectile is at 148m above the ground and its speedis 124 m/s totalenergy=Ke+Pe=1/2mv^2+mgh=52(1/2x124^2+9.8x148)=(4.7x10^5J)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 = 124cos30=107.38m/s, but dueto friction its velocity is reduced to 87.9m/s intial vertical veolocity= 124sin30=62 it is given body has reached upto 308m .from ground,as it wasthrown from cliff which is 148 above ground distance travelled byprojectile vertically from the point of throw=308-148=160m we shall calculate the real vertical speed due tofriction using y=160 m v^2-u^2=2gs; at the top V=0 so U^2=2x9.8 m/s^2x160 m=56 due to friction new velocity of projectile is 87.9i+56j magnitude= (87.9x87.9+56x56)=104.21m/s work done by air =1/2m Vi^2- 1/2 mVf^2 =1/2x52(124x124-104.21x104.21)=(1.17x10^5)answer c)work done by air in upward movemnt = 1.17x10e5 ;it is givenair has done 1.5 time more work downward; so total work done by air2.5x1.17x10e5 initial ke of body was=1/2x52x124x124 final ke= 1/2x52xV^2 final ke= intial ke - work done by friction+mgx164 as projectile reaches ground 1/2x52V^2=1/2x52x124x124-2.5x1.17x10e5+52x10x148 V=83.98m/s intially projectile is at 148m above the ground and its speedis 124 m/s totalenergy=Ke+Pe=1/2mv^2+mgh=52(1/2x124^2+9.8x148)=(4.7x10^5J)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 = 124cos30=107.38m/s, but dueto friction its velocity is reduced to 87.9m/s intial vertical veolocity= 124sin30=62 it is given body has reached upto 308m .from ground,as it wasthrown from cliff which is 148 above ground distance travelled byprojectile vertically from the point of throw=308-148=160m we shall calculate the real vertical speed due tofriction using y=160 m v^2-u^2=2gs; at the top V=0 so U^2=2x9.8 m/s^2x160 m=56 due to friction new velocity of projectile is 87.9i+56j magnitude= (87.9x87.9+56x56)=104.21m/s work done by air =1/2m Vi^2- 1/2 mVf^2 =1/2x52(124x124-104.21x104.21)=(1.17x10^5)answer c)work done by air in upward movemnt = 1.17x10e5 ;it is givenair has done 1.5 time more work downward; so total work done by air2.5x1.17x10e5 initial ke of body was=1/2x52x124x124 final ke= 1/2x52xV^2 final ke= intial ke - work done by friction+mgx164 as projectile reaches ground 1/2x52V^2=1/2x52x124x124-2.5x1.17x10e5+52x10x148 V=83.98m/s

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