$A$ fighter plane, flying horizontally with a speed of $360 \text{ km/h}$ at an altitude of $500 \text{ m}$, drops a bomb for a target straight ahead of it on the ground. At what approximate distance ahead of the target should the bomb be dropped? Assume that acceleration due to gravity $g = 10 \text{ m/s}^2$. Neglect air drag.

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:

$A$ small object is thrown at an angle $45^{\circ}$ to the horizontal with an initial velocity $v_0$. The velocity is averaged for the first $\sqrt{2} \,s$ and the magnitude of the average velocity comes out to be the same as that of the initial velocity, i.e.,$|v_0|$. The magnitude $|v_0|$ will be (take $g=10 \,m/s^2$):

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

If the height of a projectile at a time of $2 \ s$ from the beginning of motion is $60 \ m$,then the time of flight of the projectile is (Acceleration due to gravity $= 10 \ m \ s^{-2}$) (in $s$)

$A$ ball is thrown at an angle $\theta$ and another ball is thrown at an angle $(90^\circ - \theta)$ with the horizontal from the same point with the same speed $40\,m/s$. The second ball reaches $50\,m$ higher than the first ball. Find their individual heights.