$A$ swimmer dived off a cliff with a running horizontal leap. What must his minimum speed be just as he leaves the top of the cliff so that he will miss the edge at the bottom,given that the ledge at the bottom is $2 \ m$ wide and $10 \ m$ below the top of the cliff?

An object is projected in the air with initial velocity $u$ at an angle $\theta$. The projectile motion is such that the horizontal range $R$ is maximum. Another object is projected in the air with a horizontal range half of the range of the first object. The initial velocity remains the same in both cases. The value of the angle of projection,at which the second object is projected,will be $.......$ degrees.

$A$ ball is projected with kinetic energy $E$ at an angle of $45^\circ$ to the horizontal. At the highest point during its flight,its kinetic energy will be

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 given an initial velocity of $(3 \hat{i} + 4 \hat{j}) \text{ m s}^{-1}$, where $\hat{i}$ is along the ground and $\hat{j}$ is along the vertical. Assuming $g = 10 \text{ m s}^{-2}$, if the equation of its trajectory can be written as $\frac{1}{9} [\beta x + \gamma x^2]$, then the value of $\gamma$ is

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 $y = 8t - 5t^2 \text{ m}$ and $x = 6t \text{ m}$,where $t$ is in seconds. The velocity with which the projectile is projected is (in $\text{ m/s}$)