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

Vedclass pdf generator app on play store
Vedclass iOS app on app store

For upward motion

$v=u-g t$ or $0=u-g t_{1}$

or $t_{1}=\frac{u}{g}$ $....(1)$

For downward motion

$v=u+g t_{2}$ or $v=0+g t_{2}$

As the body falls back to the earth with the same velocity, it was thrown vertically upwards.

$\therefore \quad v=u$

$u=0+g t_{2}$ or $t_{2}=\frac{u}{g}$ $...(2)$

From $(1)+(2)$ the statement is proved.

Similar Questions

Write true or false for the following statements

Motion along a curved line is called translatory or rectilinear motion.

$(a)$ Derive second equation of motion $S=u t+\frac{1}{2} a t^{2}$ graphically where the symbols have their usual meanings.

$(b)$ A car accelerates uniformly from $18\, km h ^{-1}$ to $36\, km h^{-1}$ in $5$ seconds. Calculate the acceleration and the distance covered by the car in that time.

Two graphs for motion of objects moving along a straight line are shown. State how the speed is changing with time in both cases.

A car is moving on a straight road with uniform acceleration. The following table gives the speed of the car at various instants of time.

Time $(s)$ $0$ $10$ $20$ $30$ $40$ $50$
Speed $\left(m s^{-1}\right)$ $5$ $10$ $15$ $20$ $25$ $30$

$(i)$ Draw the speed$-$time graph representing the above set of observations.

$(ii)$ Find the acceleration of the car.

A piece of stone is thrown vertically upwards. It reaches its maximum height in $3$ second. If the acceleration of the stone be $9.8\, m s ^{-2}$ directed towards the ground, calculate the initial velocity of the stone with which it is thrown upwards. Find the maximum height attained by it.