Question

A particle is thrown vertically upwards. Its velocity at half of the height is $10m/s$, then the maximum height attained by it will be: $(g = 10m/{s^2})$
A) $10m$
B) $20m$
C) $15m$
D) $25m$

Answer 1

Hint: Recall that if the velocity or the acceleration of the particle is to be explained, then the equations of motion are used. These equations of motion are used to describe behaviour of a system when expressing the motion of a particle as a function of time.

Complete step by step solution:
Let the maximum height at which the particle is thrown be $H$
Using the formula of third equation of motion,
$ \Rightarrow {v^2} = {u^2} - 2gs$---(i)
Where ‘v’ is the final velocity of the particle
‘u’ is the initial velocity
‘g’ is the acceleration due to gravity
‘s’ is the distance travelled
Given that the velocity at half the height is $v = 10m/s$
Therefore, equation (i) becomes,
$\Rightarrow {(10)^2} = {u^2} - \dfrac{{2gH}}{2}$
$ \Rightarrow {(10)^2} = {u^2} - gH$---(ii)
When the particle is at its maximum height, then the velocity will be zero.
$ \Rightarrow v = 0$
Therefore, equation (i), can be written as
$ \Rightarrow {(0)^2} = {u^2} - 2gH$
$ \Rightarrow {u^2} = 2gH$---(iii)
Substitute the value from equation (iii) to equation (ii), we get
$ \Rightarrow 100 = 2gH - gH$
$ \Rightarrow gH = 100$
Given that the value of ‘g’ is 10m/s, so above equation becomes
$ \Rightarrow H = \dfrac{{100}}{{10}}$
$ \Rightarrow H = 10m$
The maximum height attained by it will be $10m$.

Hence, Option A is the right answer.

Note: It is to be noted that when a ball is thrown in an upward direction, it has some initial velocity. But when it moves in an upward direction, then its speed decreases. This is because of the influence of force of gravity that acts in an opposite direction to the motion of the ball. Therefore, it can be said that a retardation force is acting on the ball. Due to this retardation force, after a certain height the velocity of the ball becomes zero.