A cannon fires a 10 lb projectile with an initial velocity of 58 m/s, at an angl

Question

A cannon fires a 10 lb projectile with an initial velocity of 58 m/s, at an angle of 15 degrees. Assuming that air resistance performs 1.00 kJ of work against the projectile, what is the maximum height of the projectile? What is the ratio of the maximum height obtained with neglecting air resistance to the maximum height attained with air resistance? (hw/o resistance)/(hw/ resistance) NOTE: Assume that the zero point of the potential energy is level with the ground, and the height of the cannon above the ground is negligible.

Explanation / Answer

10 lb = 4.536 kg
Kinetic energy at projection = 1/2 mv^2 = 0.5 x 4.536 x 58^2 = 7630 J
Energy lost = 1000 J

Resulting Potential energy = mgh = 7630- 1000 = 6630 J
4.536 x 9.81 x h = 6630
height, h = 6630 / 44.5 = 149 m

Assuming no losses,
h = 7630 / 44.5 = 171.46 m

Ratio : 171.46 : 149 = 1.15 : 1

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