Question

A body is projected vertically upward with velocity u. If it passes through a certain point during its upward motion after time t, then it will pass through the same point during its return journey further after time t in (Take g as acceleration due to gravity)
A. $\dfrac{2u}{g}$
B. $2\left[ \dfrac{u}{g}+t \right]$
C. $\left[ \dfrac{u}{g}-t \right]$
D. $2\left[ \dfrac{u}{g}-t \right]$

Answer 1

We need to understand what is asked in question. So, what the question says is that a body is thrown upward with initial velocity u, it reaches a certain point say A in time t. Then the body attains its maximum height say B and starts descending down due to gravity g and crosses the same point in some time say t. What we are asked is the total time of journey of object from point A to B and back to A.

Formula Used:
$v=u+at$

Complete step by step answer:
We have been given that,
Initial velocity of object = u
Acceleration due to gravity = g
and,
time taken by body to reach point A = t
So, first let us find out the total time taken by body to reach its maximum height,
It is given by formula,
$v=u+at$
Now, the body is moving in the opposite direction of gravitational force. Therefore, the acceleration due to gravity will be negative i.e. –g
After substituting,
$v=u-gt$
Let's say total time taken by body to reach its maximum height B is T
Therefore, t = T
So, we get
$v=u-gT$
Again, we know that at maximum height final velocity v = 0,
Therefore,
$0=u-gT$
On rearranging the terms,
$\Rightarrow u=gT$
$\therefore T=\dfrac{u}{g}$
Now the time taken by body to travel from A to B can be given by
$\Rightarrow T-t=\dfrac{u}{g}-t$…….. (substituting $T=\dfrac{u}{g}$)
The body travels same distance i.e. from A to B and back To A in same time
Therefore, time taken by the body to travel from A back to A will be twice the obtained time
So, the total time taken by body in its journey from A-B-A is $2\left( \dfrac{u}{g}-t \right)$.

Therefore, the correct answer is option D.

Note:
While solving the above question the acceleration due to gravity is taken as negative, when the body is travelling upwards because the datum plane is assumed to be on the initial position of the object. It would make no change in the answer if it was taken at the maximum height because then velocities would be reversed that is velocity at datum plane or max height will be the initial velocity and on the ground will be final i.e. zero.