Q. A ball is thrown vertically upwards. It was observed at a height $h$ twice with a time interval $\mathrm{\Delta}t$. The initial velocity of the ball is
A) $\sqrt{8gh + g^{2}(\mathrm{\Delta}t)^{2}}$
B) $\sqrt{8gh + \left( \frac{g\mathrm{\Delta}t}{2} \right)^{2}}$
C) $\frac{1}{2}\sqrt{8gh + g^{2}(\mathrm{\Delta}t)^{2}}$
D) $\sqrt{8gh + {4g}^{2}(\mathrm{\Delta}t)^{2}}$
Explanation
(c) Let the ball be at height $h$ at time $t$ and time $(t + \mathrm{\Delta}t).$
Then,
$h = ut - \frac{1}{2}gt^{2}$ ….(i)
and $h = u(t + \mathrm{\Delta}t) - \frac{1}{2}g(t - \mathrm{\Delta}t)^{2}$ ...(ii)
Equating Eqs. (i) and (ii), we get
$t = \frac{2u - g\mathrm{\Delta}t}{2g}$ ….(iii)
Substituting Eq. (iii) in Eq. (i), we get
$h = \frac{4u^{2} - g^{2}(\mathrm{\Delta}t)^{2}}{8g}$
$\Rightarrow \ u = \frac{1}{2}\sqrt{8gh + g^{2}(\mathrm{\Delta}t)^{2}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.