$A$ stone is thrown vertically at a speed of $30 \,ms^{-1}$ making an angle of $45^{\circ}$ with the horizontal. What is the maximum height reached by the stone (in $\,m$)? Take $g=10 \,ms^{-2}$.

$A$ ball is thrown from the location $(x_0, y_0) = (0, 0)$ of a horizontal playground with an initial speed $v_0$ at an angle $\theta_0$ from the $+x$-direction. The ball is to be hit by a stone,which is thrown at the same time from the location $(x_1, y_1) = (L, 0)$. The stone is thrown at an angle $(180^{\circ} - \theta_1)$ from the $+x$-direction with a suitable initial speed $v$. For a fixed $v_0$,when $(\theta_0, \theta_1) = (45^{\circ}, 45^{\circ})$,the stone hits the ball after time $T_1$,and when $(\theta_0, \theta_1) = (60^{\circ}, 30^{\circ})$,it hits the ball after time $T_2$. In such a case,$(T_1 / T_2)^2$ is. . . . .

The velocity of a projectile at the initial point $A$ is $(2\hat i + 3\hat j) \text{ m/s}$. Its velocity (in $\text{m/s}$) at point $B$ 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 $y = (8t - 5t^2) \ m$ and $x = 6t \ m$,where $t$ is in seconds. The angle with the horizontal at which the projectile was projected is:

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:

$A$ body is projected from the earth at an angle of $30^{\circ}$ with the horizontal with some initial velocity. If its range is $20 \ m$,the maximum height reached by it is: (in meters)