Question

Prove that, for a body thrown vertically upwards, the time of ascent is equal to the time of descent.

Answer 1

Suppose a body is projected in the upward direction with a velocity u and H is maximum height attained by the body.
The velocity of the body at maximum height will be zero, i.e. v = 0.
Acceleration due to gravity (g) acts on body in the downward direction.
From the first equation of motion, $v=u+at$
$\Rightarrow 0=u-g{t}_{ascent}$
$\Rightarrow t_{ascent}=\frac{u}{g}$
Let t be total time of flight. From the second equation of motion,
$S=ut+\frac{1}{2}a{t}^2$
$\Rightarrow 0=t(u-\frac{1}{2}gt)$
$\Rightarrow t=0,t=\frac{2u}{g}$
Now, $t_{ascent}+t_{descent}=t=\frac{2u}{g}$
$\Rightarrow t_{descent}=\frac{u}{g}$
$\Rightarrow t_{ascent}=t_{descent}=\frac{u}{g}$