$A$ 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

According to the quantum theory,a photon of electromagnetic radiation of frequency $v$ has energy $E = h v$,where $h$ is known as Planck's constant. According to the theory of relativity,a particle of mass $m$ has equivalent energy $E = m c^2$,where $c$ is the speed of light. Thus,a photon can be treated as a particle having effective mass $m = \frac{h v}{c^2}$. If a flash of light is sent horizontally in the Earth's gravitational field,then photons while travelling a horizontal distance $d$ would fall through a vertical distance given by:

$A$ particle aimed at a target, projected with an angle $15^{\circ}$ with the horizontal is short of the target by $10 \ m$. If projected with an angle of $45^{\circ}$ it is away from the target by $10 \ m$, then the angle of projection to hit the target is

$A$ projectile is thrown with velocity $u$ making an angle $\theta$ with the vertical. It just crosses the tops of two poles each of height $h$ after $1\,s$ and $3\,s$,respectively. The maximum height of the projectile is ............ $m$.

$A$ stone is projected horizontally with a speed $10 \ m/s$ from an $80 \ m$ high building. The distance of the target on the ground from the foot of the building is $.... \ m$ $(g = 10 \ m/s^2)$.

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