Question

A body is projected,vertically upward,with a certain initial velocity.show that its time of ascent and that of descent are equal to each other .

Answer 1

1. let us consider a body thrown from A (on earth), with initial velocity u in a vertically upward direction.
2. Its velocity decreases due to ‘g’ acting in the opposite direction till it comes to rest at a height h.
3. Let 't' be the time taken to reach the highest point B.
4. Due to ‘g’ the body starts moving back from B to.

maximum height (h) attained by the body.

Initial velocity at A= +u.

Final velocity at B= 0.

Acceleration a= - g.

Displacement S= +h =?

Apply relation now.

$$\begin{gathered} v^2-u^2=2as \\ 0-u^2=2(-g)\left(h\right). \\ -u^2=-2gh. \\ h=\frac{u^2}{2g}\to equation1. \end{gathered}$$

This is the height reached by the body.

(i) Time as ascent ‘t’ (from A to B ).

Initial velocity at A=+u.

Final velocity at B=0.

Acceleration , a = - g.

Time ,t= ?

Apply the relation, v=u+at.

$$\begin{gathered} 0=u-gt. \\ u=gt. \\ t=\frac{u}{g}. \end{gathered}$$

(ii) Time of descent ‘t’ '(from B to A)

Initial velocity at B =0 .

Acceleration ,a= - g (downward )

Displacement (B to A) , S= -h (downward)

Time (from B to A), t' = ?

Applying the relation ,$S=ut+\frac{1}{2}\times a\times t^2.$

$$\begin{gathered} -h=0\times t\text{'}\times \frac{1}{2}(-g)t{\text{'}}^2. \\ -h=-\frac{1}{2}gt{\text{'}}^2. \\ t{\text{'}}^2=\frac{2h}{g}ort\text{'}=\sqrt{\frac{2h}{g}.} \end{gathered}$$

Substitute fro h from equation 1.

$t\text{'}=\sqrt{\frac{2}{g}\times \frac{u^2}{2g}}=\sqrt{\frac{u^2}{g^2}}.t\text{'}=\pm \frac{u}{g}.$

We can reject negative sign .being meaningless .

$t\text{'}=\frac{u}{g}.$

It is clear that time of ascent =time of descent are equal .