A 48.0-kg projectile is fired at an angle of 30.0° above the horizontal with an

Question

A 48.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 1.40 102 m/s from the top of a cliff 132 m above level ground, where the ground is taken to be y = 0.
(a) What is the initial total mechanical energy of the projectile?
J

(b) Suppose the projectile is traveling 99.2 m/s at its maximum height of y = 336 m. How much work has been done on the projectile by air friction?
J

(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?
m/s

Explanation / Answer

The total mech. energy would be the sum of the kinetic energy and potential energy of the body.

M.E.= 0.5 mv2 + mgh

taking v=102m/s as its not clear in the ques (initial speed of 1.40 102 m/s)

0.5 * 48 * 1022 + 48*10*132 = 313056J

part b is wrong as at the top the velocity will only be in x direction which shouldn't be greater than 51m/s(102sin(30))

part c is not clear

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