Question

A body thrown vertically up from the ground passes the height $10.2\;{\text{m}}$ twice at an interval of $10s$ . What was its initial velocity? (in ${\text{m}}/{\text{s}})$
(A) $52$
(B) $53$
(C) $51$
(D) $49$

Answer 1

In the question, it is said that the vertically moving body passes the particular given height twice in $10$ seconds. Then it is clear that it will go up for $5$ seconds and the body will spend the next $5$ seconds to return down.
Formula Used: We will use the basic formulas of motion to solve this question. These are the following formulas to be used:
$v = u - gt$
$s = ut - \dfrac{1}{2}a{t^2}$
${H_{\max }} = \dfrac{{{u^2}}}{{2g}}$
Where
$v$ is the final velocity of the body
$u$ is the initial velocity of the body
$s$ is the displacement of the body
$a$ is the acceleration of the body
$g$ is the acceleration due to gravity acting on the body
$t$ is the time taken by the body
${H_{\max }}$ is the maximum height attained by the body

Complete solution Step-by-Step:
It is provided to us in the question that $g = 10m/{s^2}$
And when the body reaches the maximum height, that is ${H_{\max }}$ , then the speed of the body will be zero. That is $v = 0m/s$
Since we know that
$v = u - gt$
Now we will substitute all the known variables into this equation
$\Rightarrow 0 = u - gt$
$\therefore u = gt$
Now, let us suppose the distance between ${H_{\max }}$ and height $10.2m$ be $x$
Then, we will use the formula $s = ut - \dfrac{1}{2}a{t^2}$
In the above equation, our assumed value $x$ denotes the distance $s$
Substituting the known variables, we get
$\Rightarrow s = 50 \times 5 - \dfrac{1}{2} \times 5 \times {5^2}$
On solving, we get
$\therefore x = 125m$
So, the maximum height ${H_{\max }} = 125 + 10.2 = 135.2m$
According to our formula,
${H_{\max }} = \dfrac{{{u^2}}}{{2g}}$
Now we will substitute the known variables into the above formula to calculate the initial speed, $u$
$135.2 = \dfrac{{{u^2}}}{{2 \times 10}}$
Therefore, upon further solving for $u$ , we get
$\therefore u = 52m/s$

So, the correct option is (A.)

Note: The basic concept of an object's motion, such as the position, velocity or acceleration of an object at different times, is described by equations of motion in kinematics. The motion of an object in $1D$ , $2D$ and $3D$ is governed by these three equations of motion. One of the most important topics in physics is the derivation of equations of motion. In this article, we will show you how, by graphical method, algebraic method and calculus method, we derive the first, second and third equations of motion.