Work done in projectile motion

A projectile is shot at some inclination to the ground. It falls at another point having R distance from the initial point on the ground. Is there any work done?

If initial velocity vector is $(u\cos A,u\sin A)$, when it hits the ground, its velocity will be $(u\cos A,-u\sin A)$. Change in vertical component of velocity is $2u\sin A$ in downward direction.

When velocity changes in direction or magnitude, a force is involved. As force is change in momentum wrt time.

Is there any work done? because as the initial total energy of the projectile and final total energy of the projectile same. Shouldn't the work done during upward motion cancel the work done in downward motion?

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Hi Thale. Just to mind you that, we're using MathJax markups ;-) –  Crazy Buddy Jan 30 at 6:24

The work-energy theorem in classical mechanics says that the change in kinetic energy of a system equals the net work done by external forces on the system. Since in this case the kinetic energy of a projectile (assuming no air resistance) is the same just after it's shot and just before it hits the ground again, the net work done by gravity (the only external force on the projectile) during its flight is zero.

The work done by gravity during the upward motion of the projectile does indeed cancel the work done during its downward motion as you have indicated. To show this is true mathematically, note that the force due to gravity always points downward, but the vertical component of the object's displacement vector is in the opposite direction on the way down, so by symmetry we see that the work done on the way up has the same magnitude, but opposite sign, as the work done on the way down, and the two cancel.

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First of all,work done is always defined as done by a force so the phrase "work done" does not stand by itself.

Secondly the magnitude of the velocity has not changed which implies that the kinetic energy has not changed which implies no work has been done by gravity. In simple words,yes,"you require a force to change the velocity but you require work done by some force to change the kinetic energy."

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