Question

A 40.0-kg projectile is fired at an angle of 30.0° above the horizontal with an initial speed of 134 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? (Give your answer to at least three significant figures.) J (b) Suppose the projectile is traveling 95.0 m/s at its maximum height of y = 319 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

Answer 1

(a) The initial total mechanical energy of the projectile is the sum of its initial kinetic and potential energy.

KE = ½ * 40 * 134^2 = 359120 J
PE = 40 * 9.8 * 132 = 51744 J
Total = 410864 J


(b) Let’s determine the kinetic and potential energy at this position.
KE = ½ * 40* 95^2 = 180500 J
PE = 40 * 9.8 * 319 = 125048 J
Total = 305548 J

If there is no air friction, the projectile’s vertical velocity at its maximum height would be 0 m/s. The only velocity at this position is its initial horizontal velocity.

Initial horizontal velocity = 134 * cos 30 = 116 m/s
Let’s determine the kinetic energy.

KE = ½ * 40 * (134 * cos 30)^2 = 269340 J

The air friction caused the projectile’s kinetic energy to decrease from 269340 J to 180500 J. The work that has been done on the projectile by air friction is equal to the decrease of its kinetic energy.

Work = 269340 – 180500 = 88840 J

(c) Work = 1.5 * 88840 = 133260 J

When the projectile hits the ground, its potential energy is 0 J. This means all of its energy at the maximum height would be converted into kinetic energy. To determine its kinetic energy at the ground, subtract the work from the total energy at the maximum height.

KE = 305548 – 133260 = 172288 J
½ * 40 * v^2 = 172288
v = 92.8 m/s.