Question

A particle when thrown. moves such that it passes from same height at $2$ and $10s$, the height is:
A. $g$
B. $2g$
C. $5g$
D. $10g$

Answer 1

Hint: In order to solve this question, we will find the distance covered by both the bodies in a given interval of time and then we will solve for the given same height and here we will use the general Newton’s equations of motion.

Formula used:
Equation of motion is
$S = ut + \dfrac{1}{2}a{t^2}$
where,
S is the distance covered by the body
u is the initial velocity of the body
a is the acceleration of the body and when body motion is under effect of gravity then $a = g$ called acceleration due to gravity.
t is the time taken by the body.

Complete step by step solution:
According to the question, let us suppose the height is h at which particle passes twice at the time of $t = 2,t = 10s$ let its initial velocity be u and acceleration due to gravity is $g$ and due to upward motion acceleration of the body will be $a = - g$. So using equation of motion $S = ut + \dfrac{1}{2}a{t^2}$ we get, at $t = 2s$
$h = u(2) - \dfrac{1}{2}g{(2)^2} \\
\Rightarrow h = 2u - 2g \to (i) \\ $
At, $t = 10s$ we have,
$h = u(10) - \dfrac{1}{2}g{(10)^2} \\
\Rightarrow h = 10u - 50g \to (ii) \\ $
Now, applying the operation five time the equation (i) and subtract the equation (ii) as $5(i) - (ii)$ we get
$5h - h = 40g \\
\Rightarrow h = 10g \\ $
So, the height at which the particle passes twice in a given period of time is $10g$.

Hence, the correct answer is option D.

Note: While solving such problems, don’t make the mistake of using the sign used for acceleration due to gravity because as per standard sign conventions we use acceleration due to gravity as negative for upward motion while positive for downward motion.