Question

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 s
B. 1.5 s
C. 1 s
D. 0.5 s

Answer 1

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.

Complete step by step answer:
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.