Q. A projectile moves from the ground such that its horizontal displacement is $x = Kt$ and vertical displacement is $y = Kt(1 - \alpha t)$, where $K$ and $\alpha$ are constants and $t$ is time. Find out total time of flight (T) and maximum height attained $(Y_{\max})$ its
A) $T = \alpha,\ Y_{\max} = \frac{K}{2\alpha}$
B) $T = \frac{1}{\alpha},\ Y_{\max} = \frac{2K}{\alpha}$
C) $T = \frac{1}{\alpha},\ Y_{\max} = \frac{K}{6\alpha}$
D) $T = \frac{1}{\alpha},\ Y_{\max} = \frac{K}{4\alpha}$
Explanation
(d) During time of flight vertical displacement becomes zero
$i.e.,$ $y = 0$
$KT(1 - \alpha T) = 0 \Rightarrow T = \frac{1}{\alpha}$
$\therefore$ Time taken by particle to attain max, height $t = T/2 =$
$\frac{1}{2\alpha};\ \therefore Y_{\max} = K.\frac{1}{2\alpha}\left( 1 - \alpha.\frac{1}{2\alpha} \right) \Rightarrow \ = \frac{K}{4\alpha}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.