projectile---physics

This question could be answered simply using the Law of Conservation of Energy.

Because the firing point and landing point are at the same level, there is no change of potential energy of the projectile.

Assume no air resistance, by conservation of energy, the kinetic energy at the firing point and landing point must be equal.

That is, (1/2)mu^2 = (1/2)m.v^2
where m is the mass of the projectile
u is the firing speed, and v is the landing speed

It is thus clear that u must equals to v

For a projectile, the velocity can always be ressolved into two perpendicular components.

In the horizontal motion, it is always moving with uniform horizontal speed. Let it be u.

Hence, the initial horizontal speed

= final horizontal speed = u

2008-10-01 20:32:19 補充：
Now, for vertical motion, let upward direction be positive

Initial speed = v

Acceleration due to gravity = -g

Vertical displacement, s = 0

By w2 = v2 - 2gs

w = -v (The vertical velocity just before striking the ground is pointing downward)

2008-10-01 20:32:26 補充：
Hence, speed just leaves the ground,

u' = (u2 + v2)1/2

Speed just before striking the ground,

u" = (u2 + w2)1/2 = (u2 + (-v)2)1/2 = (u2 + v2)1/2

Hence, the two speeds are the same.