Question

A ball is thrown vertically upwards. It was observed at a height h twice with a time interval $\Delta t$. The initial velocity of the ball is
A. $\sqrt {8gh + {{(g\Delta t)}^2}} $
B. $\sqrt {2gh + {{(\dfrac{{g\Delta t}}{2})}^2}} $
C. $\sqrt {8gh + {{(2g\Delta t)}^2}} $
D. $\sqrt {2gh} $

Answer 1

Hint:n order to solve this question, we will use newton’s equations of motion to derive the relation between height and time taken and then we will solve for the initial speed of the ball in terms of time interval and height attained by the ball.

Formula used:
One of the Newton’s 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, we have given that distance covered by the ball is height H and let its initial velocity be ‘u’ and time taken to reach at this height is t and acceleration due to gravity will be $ - g$ due to upward motion so, using equation of motion $S = ut + \dfrac{1}{2}a{t^2}$ and putting the values we get;
$H = ut - \dfrac{1}{2}g{t^2} \\
\Rightarrow {t^2} - \dfrac{{2u}}{g}t + \dfrac{{2h}}{g} = 0 \\ $
So, we have a quadratic equation in terms of time ‘t’ so there will be two possible values of ‘t’ and roots of quadratic equation are related as
${t_1} + {t_2} = \dfrac{{2u}}{g} \\
\Rightarrow {t_1}{t_2} = \dfrac{{2h}}{g} \\ $
From these relation we can find the term $\Delta t = {t_2} - {t_1}$ as
${({t_2} - {t_1})^2} = {({t_2} + {t_1})^2} - 4{t_1}{t_2}$
On putting the values of required parameters we get,
$ \Delta {t^2} = {(\dfrac{{2u}}{g})^2} - 4(\dfrac{{2h}}{g}) \\
\Rightarrow {(\dfrac{{2u}}{g})^2} = \Delta {t^2} + 4(\dfrac{{2h}}{g}) \\ $
Solving for u we get,
$u = \sqrt {2gh + {{(\dfrac{{g\Delta t}}{2})}^2}} $

Hence, the correct answer is option B.

Note: It should be remembered that, other two equations of motion are also very useful in such kinematics question and they are $v = u + at$ where v is the final velocity of the body and other one is ${v^2} - {u^2} = 2aS$.