If the velocity at the maximum height of a projectile projected at an angle of $45^{\circ}$ is $20 \,m \,s^{-1}$, then the maximum height reached by the projectile is (Acceleration due to gravity $=10 \,m \,s^{-2}$ ) (in $\,m$)

$A$ cricket player can throw a ball with an initial speed of $30 \,m/s$. What is the maximum range the player can throw the ball? Neglect air resistance. [Take $g = 10 \,m/s^2$]

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:

Two bullets are fired horizontally and simultaneously towards each other from the rooftops of two buildings $100 \; m$ apart and of the same height of $200 \; m$ with the same velocity of $25 \; m/s$. When and where will the two bullets collide? $(g = 10 \; m/s^2)$

$A$ body is projected with a velocity $(\hat{i} + 2\hat{j}) \text{ ms}^{-1}$,where $\hat{i}$ is along the horizontal and $\hat{j}$ is vertically upward. Then the equation of its trajectory is $(g = 10 \text{ ms}^{-2})$.

$A$ particle of mass $m$ is thrown with an initial speed $u$ at an angle $\theta$ with the horizontal. Find the torque of its weight about the point of projection when it reaches the highest point.