Question

Show that the speed with which a projectile leaves the ground is equal to its speed just before it strikes the ground at the end of its journey, assumilng the firing level equals the landing level.

Answer 1

Answer:

Thus, the velocity at the time of strike is same as the velocity at the time of projection.

Explanation:

Let a projectile is projected vertically upwards with a speed of u and reaches to the maximum height H.

At maximum height , the speed is zero and then the projective comes back on the ground.

Use the third equation of motion

$v^2 = u^2 + 2 g h \\\\0 = u^2 - 2 g H\\\\\u =\sqrt{2gH}$

Now let the velocity at the time of strike is v'.

Use third equation of motion, here initial velocity is zero.  

$v'^2 = 0 + 2 g H \\\\v = \sqrt{2gH}$

Thus, the velocity at the time of strike is same as the velocity at the time of projection.