The horizontal and vertical displacements $x$ and $y$ of a projectile at a given time $t$ are given by $x = 6t \text{ m}$ and $y = 8t - 5t^2 \text{ m}$. The range of the projectile in metres is:

$A$ projectile is thrown with velocity $v$ at an angle $\theta$ with the horizontal. When the projectile is at a height equal to half of the maximum height,the vertical component of the velocity of the projectile is ...........

$A$ particle moves in a plane with constant acceleration in a direction different from the initial velocity. The path of the particle will be

If the velocity at the maximum height of a projectile projected at an angle of $45^{\circ}$ is $20 \,m \,s^{-1}$, then the maximum height reached by the projectile is (Acceleration due to gravity $=10 \,m \,s^{-2}$ ) (in $\,m$)

For a given velocity,a projectile has the same range $R$ for two angles of projection. If $t_1$ and $t_2$ are the times of flight in the two cases,then:

$A$ body is projected from the ground obliquely. During its downward motion,the power delivered by gravity to it: