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.

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

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

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

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

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the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s