Q. The height $y$ and the distance $x$ along the horizontal plane of a projectile on a certain planet (with no surrounding atmosphere) are given by $y = 8t - {5t}^{2}$ metre and $x = 6t$ metre, where $t$ is in second. The velocity with which the projectile is projected, is
A) ${14\ ms}^{- 1}$
B) ${10\ ms}^{- 1}$
C) ${8\ ms}^{- 1}$
D) ${6\ ms}^{- 1}$
Explanation
(b) $x = (ucos\theta)t = 6$
$ucos\theta = \frac{x}{t} = 6$
$y = (usin\theta)t = - \frac{1}{2}gt^{2}$
$y = 8t - {5t}^{2} \Longrightarrow usin\theta = 8$
$u = 10m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.