A projectile is given an initial velocity of $(\hat i+2\hat j)\,m/ s$ where $\hat i$ is along the ground and $\hat j$ is along the vertical. If $g = 10\,m/s^2,$ the equation of its trajectory is
$y = x -5x^2$
$y =2 x -5x^2$
$4y = 2x -5x^2$
$4y = 2x -25x^2$
A particle is projected with a velocity $v$ such that its range on the horizontal plane is twice the greatest height attained by it. The range of the projectile is (where $g$ is acceleration due to gravity)
Average velocity of a particle is projectile motion between its starting point and the highest point of its trajectory is : (projection speed = $u$, angle of projection from horizontal= $\theta$)
A projectile is thrown into space so as to have maximum horizontal range $R$. Taking the point of projection as origin, the coordinates of the points where the speed of the particle is minimum are-
A point $P$ moves in counter clock wise direction on a circular path as shown in figure. The movement of $'P'$ is such that it sweeps out a length $S = t^3 + 5$, where $'S'$ is in meter and $t$ is in seconds. The radius of the path is $20\, m$. The acceleration of $'P'$ when $t = 2\, sec$. is nearly ......... $m/s^2$