Question

If a body is projected vertically up it's velocity decreases to half of its initial velocity at a height 'h' above the ground The maximum height reached by it is?

Answer 1

In order to answer the given question to know the maximum height we will be using the velocity acceleration equation that is Initial velocity $\left( u \right)$ squared plus two times acceleration $\left( a \right)$ times displacement $\left( s \right)$ equals final velocity $\left( v \right)$ squared.
For equations using the Earth's gravitational force as the acceleration rate of an object, use standard gravity, $a = 9.80665{\text{ }}m{s^{ - 2}}$ .

Complete step by step answer:
Let us consider the initial velocity be $= u\,m{s^{ - 1}}$
The acceleration due to gravity is $g = 9.8m{s^{ - 2}}$
Resolving it in a positive manner as it is in the upward direction $= + ve$
The velocity at height $h = \dfrac{u}{2}m{s^{ - 1}}$
Using the equation of motion to solve a problem
${v^2} = {u^2} + 2as$
Putting the value of $v$ in the equation
${\left( {\dfrac{u}{2}} \right)^2} = {u^2} - 2gh \\\ = \dfrac{{{u^2}}}{4} = {u^2} - 2gh \\\$
So, from here, value of $2gh$
$\Rightarrow 2gh = {u^2} - \dfrac{{{u^2}}}{4} \\\ \Rightarrow 2gh = \dfrac{{4{u^2} - {u^2}}}{4} = \dfrac{{3{u^2}}}{4} \\\$
From here value of ${u^2}$ will be
${u^2} = \dfrac{8}{3}gh\,..........\left( 1 \right)$
Let at the maximum height $v = 0$
Let the maximum height be $= y\,m$
Therefore, $0 = {u^2} - 2gy$
Plugging in the value of equation (1)
$\dfrac{8}{3}gh = 2gy \\\ \therefore y = \dfrac{4}{3}h \\\$
Therefore, the maximum height is $\dfrac{4}{3}h$.

Additional Information:-
This equation is valid while moving in a straight line. Because the distance between the initial point and the final point in a straight line motion is equal to displacement, we may conclude that $S$ is speed in this situation. However, in this case, $S$ stands for displacement.
If we return to the starting place and take a step in the opposite direction, displacement will be negative. So whether $S$ is negative or positive depends on the condition, but it is evident that $S$ is displacement.

Note: It should be noted that using kinematic equations to determine the height obtained by a body thrown upwards at any velocity is not valid. Kinematic equations are only valid if the acceleration is uniform. Only at the earth's surface does gravity's acceleration become uniform.