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# A ball is thrown vertically upward attains a maximum height of 45 m. The time after which the velocity of the ball becomes equal to half the velocity of projection (use $g = 10{\text{ m/}}{{\text{s}}^2}$)A. 2 sB. 1.5 sC. 1 sD. 0.5 s

Last updated date: 13th Jun 2024
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Hint: Projectile motion is a form of motion experienced by an object that is projected near the Earth’s surface and moves along a curved path under the action of gravity.
In this question, height and final velocity are given; hence we will find the velocity of projection, then using this velocity of projection, we will find the time.

Given the maximum height reached by the ball h=45m
Let the velocity of projection be u and the final velocity at the top height $v = 0$
We know newton law of motion is given as
${v^2} = {u^2} - 2gh - - (i)$[ $- g$since the ball is moving in opposition to the direction of gravity]
Here since the final velocity$v = 0$, so we can write equation (i) as
$0 = {u^2} - 2\left( {10} \right)\left( {45} \right) \\ {u^2} = 900 \\ u = 30{\text{ m/sec}} \\$
So the velocity of the velocity is $u = 30{\text{ m/sec}}$
Now we need to find the time after which the velocity of the ball becomes equal to half the velocity of projection i.e.
$v = \dfrac{{30}}{2} = 15{\text{ m/sec}}$
Now use the newton law of motion to find the time which is given as
$v = u - gt - - (ii)$
So by substituting the value of velocity, we get
$15 = 30 - 10t \\ 10t = 15 \\ t = 1.5s \\$
Therefore we can say time after which the velocity of the ball becomes equal to half the velocity of projection is$t = 1.5s$
Option (B) is correct.

Note:Students must know that whenever any object or a particle is moving against the velocity, then its gravity will act in the opposite direction. The projectile has a single force that acts upon it, which is the force of gravity.