Question

What is the time of descent for a body projected vertically upward with velocity $v$ ?

• A. Same as time of ascent
• B. $\frac{v}{g}$
• C. $\frac{u}{g}$ , u = final velocity
• D. Both A and B

Answer 1

Answer: (D)

Both A and B

Ascent:

$v=u-g{t}_a$

Or,

$0=u-g{t}_a$

Or,

Time of ascent, $t_a=u/g$

Now, $v^2=u^2-2gh$

Or,

$u=\sqrt{2gh}$

or

$t_a=\sqrt{2gh}/g$ ---(1)

Descent:
$v^2=u^2+2gh$

here $u=0$
So, $v=$$\sqrt{2gh}$
Now applying $v=u+g{t}_d$ and putting $u=0$,

we get $v=g{t}_d$

So time of descent,$t_d=\frac{v}{g}=\frac{\sqrt{2gh}}{g}$ ---(2)

From (1) and (2) we get,

$t_a=t_d$