A point moves in $x-y$ plane as per $x=kt,$ $y = kt\left( {1 - \alpha t} \right)$ where $k\,\& \,\alpha \,$ are $+ve$ constants. The equation of trajectory is
$y = x - \frac{{\alpha {x^2}}}{k}$
$y = x + \frac{{\alpha {x^2}}}{k}$
$x = y - \frac{{\alpha {y^2}}}{k}$
$x = y - \frac{{\alpha {y^2}}}{k}$
The equation of a projectile is $y=a x-b x^2$. Its horizontal range is ......
A particle initially at rest is subjected to two forces. One is constant, the other is a retarding force proportional to the particle velocity. In the subsequent motion of the particle :
The coordinates of a particle moving in a plane are given by $x = a\cos (pt)$ and $y(t) = b\sin (pt)$ where $a,\,\,b\,( < a)$ and $p$ are positive constants of appropriate dimensions. Then
A body starts from rest from the origin with an acceleration of $6 \;m / s^2$ along the $x$-axis and $8\; m / s^2$ along the $y$-axis. Its distance from the origin after $4\; seconds$ will be
If the initial velocity in horizontal direction of a projectile is unit vector $\hat{i}$ and the equation of trajectory is $y =5 x (1- x )$. The $y$ component vector of the initial velocity is.
(Take $g=10\,m / s ^{2}$ )