A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an

Question

A 54.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 1.26 102 m/s from the top of a cliff 128 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 89.3 m/s at its maximum height of y = 293 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

a)The total energy will be the sum of the kinetic and potential due to gravity.
TE=PE+KE
TE=mgh+(1/2)mv2=54(9.8)(128)+(1/2)(54)(126)2

TE1=496389.6 Joules (497 KJ)

b)We once again find the total energy, but at this new spot

TE=PE+KE
TE=mgh+(1/2)mv2=54(9.8)(293)+(1/2)(54)(89.3)2

TE2=370366.83 Joules (370 KJ)

Now to find the work we take the difference in the total mechanical energies of the two spots

W=TE1+TE2= 496389.6-370366.83

W=126022.77 Joules (126KJ)

c) Now we should figure out what the total amount of work will be.

WT=W+(1.5)W=2.5W=2.5(126022.77)

WT=315053.925 joules

Now we say that all the energy an instant before the object hits the ground will be purely kinetic. This energy will be equal to the initial energy minus the worked done on it by air friction

KE=TE1-WT=496389.6-315053.925

KE=181335.675

(1/2)(54)v2=181335.675

v=81.95 m/s

Hope that helps

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