Q. A body is thrown vertically upwards with velocity $u.$ The distance travelled by it in the fifth and the sixth seconds are equal. The velocity $u$ is given by $\left( g = 9.8\ {ms}^{- 2} \right)$
A) $24.5\ {ms}^{- 1}$
B) $49.0\ {ms}^{- 1}$
C) $73.5\ {ms}^{- 1}$
D) $98.0\ {ms}^{- 1}$
Explanation
(b) The distance travelled in $t$ sec in upward motion is
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$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ s = u - \frac{1}{2}\ g(2t - 1)$
$\therefore\ \ \ \ \ \ \ \ \ \ AB = u - \frac{1}{2}g(2 \times 5 - 1)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ AB = u - \frac{1}{2}\ 9g$
Distance travelled in 1 s in the downward direction is
$\ \ \ \ \ \ \ \ \ \ \ \ \ BA = 0 + \frac{1}{2}\ g(1)^{2}$
It is given that these distance are equal. Therefore,
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u - \frac{9g}{2} = \frac{1}{2}g$
$\Rightarrow \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ u = 5 \times 9.8 = 49\ {ms}^{- 1}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.