Question

distance covered by a body projected vertically in the 1st second of his upward journey equal to distance covered by it in the second of his entire journey?​

Answer 1

Explanation:

Let the speed of projection be u and the time taken to reach maximum height be t.

St=ut−0.5gt2

St−1=u(t−1)−0.5g(t−1)2

Solving these two equations, we get 

St−St−1=u−gt+2g 

which is the distance traveled in the last second.

For a vertically projected body, u=gt (because 't' is the time when it reachs max. height), hence

st−st−1=2g

This shows that the distance traveled by a body projected vertically up is independent of the initial speed.

So, even if the body is projected with double the speed, the body covers the same distance, d in the last second.