$A$ stone is thrown at an angle $\theta$ to the horizontal reaches a maximum height $H$. Then the time of flight of the stone will be

$A$ body of mass $10 \ kg$ is projected at an angle of $45^{\circ}$ with the horizontal. The trajectory of the body is observed to pass through a point $(20, 10)$. If $T$ is the time of flight,then its momentum vector,at time $t = \frac{T}{\sqrt{2}}$,is. [Take $g = 10 \ m/s^2$]

If the initial velocity in the horizontal direction of a projectile is the unit vector $\hat{i}$ and the equation of the trajectory is $y = 5x(1 - x)$,find the $y$-component vector of the initial velocity. (Take $g = 10\,m/s^2$) (in $,\hat{j}$)

$A$ ball is projected with kinetic energy $E$, at an angle of $60^{\circ}$ to the horizontal. The kinetic energy of this ball at the highest point of its flight will be:

$A$ particle is thrown with a speed $u$ at an angle $\theta$ with the horizontal. When the particle makes an angle $\phi$ with the horizontal,its speed changes to $v$,where

$A$ boy throws a ball upwards with velocity $v_0 = 20\, m/s$. The wind imparts a horizontal acceleration of $4\, m/s^2$ to the ball. The angle $\theta$ from the vertical at which the ball must be thrown so that the ball returns to the boy's hand is $(g = 10\, m/s^2)$