Two projectiles $A$ and $B$ are thrown with initial velocities of $40\,m / s$ and $60\,m / s$ at angles $30^{\circ}$ and $60^{\circ}$ with the horizontal respectively. The ratio of their ranges respectively is $\left( g =10\,m / s ^2\right)$

A boy playing on the roof of a $10\, m$ high building throws a ball with a speed of $10\,m/s$ at an angle of $30^o$ with the horizontal. How far from the throwing point will the ball be at the height of $10\, m$ from the ground ?  $\left[ {g = 10\,m/{s^2},\sin \,{{30}^o} = \frac{1}{2},\cos \,{{30}^o} = \frac{{\sqrt 3 }}{2}} \right]$

A projectile $A$ is thrown at an angle $30^{\circ}$ to the horizontal from point $P$. At the same time another projectile $B$ is thrown with velocity $v_2$ upwards from the point $Q$ vertically below the highest point $A$ would reach. For $B$ to collide with $A$, the ratio $\frac{v_2}{v_1}$ should be

If the initial velocity of a projectile be doubled, keeping the angle of projection same, the maximum height reached by it will

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 first object. The initial velocity remains same in both the case. The value of the angle of projection, at which the second object is projected, will be $.......$ degree.

Show that for a projectile the angle between the velocity and the $x$ -axis as a function of time is given by

$\theta(t)=\tan ^{-1}\left(\frac{v_{0 y}-g t}{v_{0 x}}\right)$

Show that the projection angle $\theta_{0}$ for a projectile launched from the origin is given by

$\theta_{0}=\tan ^{-1}\left(\frac{4 h_{m}}{R}\right)$

Where the symbols have their usual meaning.