Question

Time of ascent is equal to
(A) Time of journey
(B) Time of descent
(C) Both (a) and (b) above
(D) None of these

Answer 1

Hint For a body thrown vertically upward, the path taking during ascent is equal to the path taken during descent. Also the velocity at its maximum height is zero, and, when air resistance is negligible, the velocity at the start is equal to the velocity at the end.

Formula used: In the solution we will be using the following formula,
$\Rightarrow v = u \pm gt$ where is the final velocity at time $t$ and is the initial velocity at time $t = 0$ , while $g$ is equal to acceleration due to gravity.

Complete step by step answer
When a body in a gravitational field is thrown vertically upward, it decelerates gradually until it reaches a velocity of zero, it hangs for a moment, then descends to the ground accelerating gradually until it gets to the maximum velocity just before it collides with the floor.
Let’s consider a body thrown upward with an initial velocity $u$ , the final velocity during ascent is $v = 0$ . The equation is given by
$\Rightarrow v = u - gt$ (assuming upward is positive and downward is negative)
With $v = 0$ we can substitute it in the equation and get,
$\Rightarrow 0 = u - g{t_1}$
On arranging we get,
$\Rightarrow u = g{t_1}$
Making $t$ subject of the formula, we have
$\Rightarrow {t_1} = \dfrac{u}{g}$
During descent, the final velocity is equal to the initial velocity when thrown, thus, $v = u$ .
Hence, from $v = u - gt$ , replacing $v$ with $- u$ , we have
$\Rightarrow - u = 0 - g{t_2}$ (it is $- u$ because velocity is now downwards)
Similarly we get the value as,
$\Rightarrow {t_2} = \dfrac{u}{g}$
Thus we get the time of ascent and descent , ${t_1} = {t_2}$
Hence time of ascent is equal to time of descent.
Thus, the correct option is B.

Note
Alternately, in general, for a projectile fired at an angle $\theta$ to the horizontal, the time of journey is given as time taken to reach maximum height plus time of descent. Mathematically,
$\Rightarrow T = {t_1} + {t_2}$
But time of journey is given as
$\Rightarrow T = \dfrac{{2u\sin \theta }}{g}$
And time taken to reach maximum height, which is time of ascent is given as
$\Rightarrow {t_1} = \dfrac{{u\sin \theta }}{g}$
Thus,
$\Rightarrow {t_2} = \dfrac{{2u\sin \theta }}{g} - \dfrac{{u\sin \theta }}{g} = \dfrac{{u\sin \theta }}{g}$
Hence, ${t_1} = {t_2}$
For a vertically thrown body $\theta = 90^\circ$ .