An object with a mass of 58.0 kg is fired with an initial speed of 1.36 times 10

Question

An object with a mass of 58.0 kg is fired with an initial speed of 1.36 times 10^2 m/s at an angle of 30.0 degree above the horizontal from a 132-m-high cliff (take ground level to be y = 0 m). Determine the initial total mechanical energy of the projectile. ] If when the projectile is at its maximum height of y = 325 m, it is traveling 96.4 m/s, determine the amount of work that has been done on the projectile by air resistance. ] What is the speed of the projectile immediately before it hits the ground if air resistance 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? m/s

Explanation / Answer

(1) total mechanical energy = Potential energy + kinetic energy
   = mgh + (1/2)mV2
h is the initial height of the projectile = 132 m
= 58*9.81*132 + (1/2)*58*(136)2 = 611489.36 J
(2) Now when the projectile reached at its maximum height
= (1/2)mV2 + mgh + WA
where WA is the work done by air resistance
= (1/2)58*(96.4)2 + 58*9.81*325 + WA = 454414.34 + WA
applying energy conservation
611489.36 = 454414.34 + WA
WA = 157075.2 J
(3) When projectile hit the ground
Energy = (1/2)mV2 + wA
its potential energy must be zero at ground level ,
And the wA = 1.5WA
Now applying energy conservation
611489.36 = (1/2)mV2 + wA
611489.36 = (1/2)mV2 + 1.5WA
V = 113.84 m/s
this will be the velocity when projectile hit the ground.

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