Question

10. a body is projected vertically up with
velocity 'u' from ground and in the last
second of its total journey distance
travelled by it is 'h'. if that body is
projected with twice the initial velocity as
that of the previous case, distance
travelled by it in the last second of its total
motion would be

Answer 1

Answer:

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

S

t

​

=ut−0.5gt

2

S

t−1

​

=u(t−1)−0.5g(t−1)

2

Solving these two equations, we get

S

t

​

−S

t−1

​

=u−gt+

2

g

​

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

s

t

​

−s

t−1

​

=

2

g

​

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

Explanation: