$A$ ball is projected at an angle $45^{\circ}$ with the horizontal. It passes through a wall of height $h$ at a horizontal distance $d_1$ from the point of projection and strikes the ground at a horizontal distance $(d_1 + d_2)$ from the point of projection. Then $h$ is:

$A$ projectile thrown from the ground has initial speed $u$ and its direction makes an angle $\theta$ with the horizontal. If at maximum height from the ground,the speed of the projectile is half its initial speed of projection,then the maximum height reached by the projectile is:
$[g = \text{acceleration due to gravity}, \sin 30^{\circ} = \cos 60^{\circ} = 0.5, \cos 30^{\circ} = \sin 60^{\circ} = \sqrt{3}/2]$

$A$ body is projected at $t = 0$ with a velocity $10\,ms^{-1}$ at an angle of $60^\circ$ with the horizontal. The radius of curvature of its trajectory at $t = 1\,s$ is $R$. Neglecting air resistance and taking acceleration due to gravity $g = 10\,ms^{-2}$,the value of $R$ is ........ $m$.

It is possible to project a particle with a given velocity in two possible ways so as to make them pass through a point $P$ at a horizontal distance $r$ from the point of projection. If $t_1$ and $t_2$ are times taken to reach this point in two possible ways,then the product $t_1 t_2$ is proportional to

$A$ 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 $(g = 10\, m/s^2)$

For a projectile,the ratio of the maximum height reached to the square of the time of flight is: