Kinematics Reasonings

# A body thrown vertically upwards takes ‘t’ time to reach the highest point. What would be the time required to reach the same point during its return?

Let $h$ be the height reached by the body in time $t$ when thrown vertically upwards by a velocity $u$. Then, $v=u-gt$ $0=u-gt$ $\therefore t = \frac{u}{g}$

and, $v^2 = u^2 – 2gh$ $0 = u^2 – 2gh$ $\therefore h = \frac{u^2}{2g}$

When the body comes down, it has to travel down the same distance. $\therefore h’ = h = \frac{u^2}{2g}$

The body comes down with initial velocity $u’ = 0$.

$\therefore h’ = u’t+\frac{1}{2}g{t’}^2$

$h’ = 0 + \frac{1}{2}g{t’}^2$

$\therefore t’ = \sqrt{\frac{2h’}{g}} = \sqrt{\frac{2u^2}{g×2g}} = \frac{u}{g} = t$

Thus, the time required to reach the highest point and the time required to come back to the same point is same.

[The equations used above are deduced from Motion Under Gravity]

#### SIMILAR QUESTIONS

A player hits a baseball at some angle. The ball goes high up in space. The player runs and catches the ball before it hits the ground. Which of the two (the player or the ball) has greater displacement?

Can an object have velocity and acceleration in perpendicular directions?

If a particle is accelerating, it is either speeding up or speeding down. Do you agree with this statement?

A ball is thrown upward with velocity ‘u’. What will be its velocity when it returns to earth? Explain.

A body is dropped from a certain point and it travels a height ‘h’ after some time. What would be the height travelled during twice that time?

Two bodies of different masses are dropped from same height. Compare their velocities on reaching the ground and the time required.