A projectile of mass m is fired horizontally with an initial speed of v0 from a

Question

A projectile of mass m is fired horizontally with an initial speed of v0 from a height of h above a flat, desert surface. Neglecting air friction, at the instant before the projectile hits the ground, find the following in terms of m, v0, h, and g.
(a) The work done by the force of gravity on the projectile.
W =



(b) The change in kinetic energy of the projectile since it was fired.
?KE =



(c) The final kinetic energy of the projectile.
KEf =


(d) Are any of the answers changed if the initial angle is changed?

Yes
No

Explanation / Answer

a. Energy = mgh, same as work done. b. Same as the amount from part a, since no friction. The change in kinetic energy since the projectile was fired = mgh. You can see this by realizing that the kinetic energy of the projectile at height h is the same when it comes down as when it started up. So the difference in kinetic energy when the projectile is almost at the ground is the potential energy at the start. Potential energy = F*d = mad = mgh in this example. c. ke = ½mv_0² + mgh The final kinetic energy is the same as the initial kinetic energy plus the potential energy. d.no change,the energies not depend on the angle of projection(from above answers we got)

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