$A$ 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 $g=10 \,ms^{-2}$.

The horizontal range $(R)$ of a projectile is $n$ times its maximum height $(H)$. Find the angle of projection $(\theta_0)$.

$A$ projectile is thrown with an initial velocity $(a \hat{i} + b \hat{j}) \ m/s$,where $\hat{i}$ and $\hat{j}$ are unit vectors along horizontal and vertical directions respectively. If the range of the projectile is twice the maximum height reached by it,then

The ratio of minimum kinetic energies of two projectiles of the same mass is $4:1$. The ratio of the maximum heights attained by them is also $4:1$. The ratio of their ranges would be: (in $:1$)

The velocity of a projectile is $(6\hat{i} + 8\hat{j}) \ m/s$. Its horizontal range is ........ $m$.

$A$ stone is projected from the ground at $t = 0$. At the time of projection,the horizontal and vertical components of velocity are $10\, m/s$ and $20\, m/s$ respectively. Find the time at which the magnitudes of tangential and normal acceleration will be equal $(g = 10\, m/s^2)$ [neglect air friction].