$A$ ball rolls from the top of a stairway with a horizontal velocity $u \; m/s$. If the steps are $h \; m$ high and $b \; m$ wide,the ball will hit the edge of the $n^{th}$ step,if $n=$

$A$ projectile is thrown at an angle $\theta$ such that it is just able to cross a vertical wall of height $H$ at its highest point,as shown in the figure. The horizontal distance from the point of projection to the wall is $\sqrt{3} H$. The angle $\theta$ at which the projectile is thrown is given by:

If the initial velocity of a projectile is doubled,keeping the angle of projection the same,the maximum height reached by it will:

$A$ particle is projected with a velocity $v$ such that its horizontal range is twice the greatest height attained. The horizontal range is

$A$ 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:

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: