A bullet fired vertically up from the ground reaches height 40 m in its path from the ground it takes further time 2 seconds to reach the same point during the descent. The total time of flight is (g = $10m{{s}^{-2}}$)
A. 4 s
B. 3 s
C. 6 s
D. 8 s

Hint: In the above question, the difference between the time of ascend and time of descent is given to us. By using that difference in the third equation of motion, we can find the relation between u and g. And the, we can use that relation to find out total time of flight
Formula used:
For solving the given question, we will be the Newton’s third equation of motion, i.e.,
$s=ut+\dfrac{1}{2}a{{t}^{2}}$

Complete answer:
Before we start solving the given question, let us take a look at all the given parameters
a = g = $10m{{s}^{-2}}$
now, let us assume the time of ascend and descend be ${{t}_{1}}$ and ${{t}_{2}}$respectively
so, we have
 ${{t}_{2}}-{{t}_{1}}=2s$
Now, using the third equation of motion,
 $s=ut+\dfrac{1}{2}a{{t}^{2}}$
On solving the quadratic equation for t, we have two solutions ${{t}_{1}}$ and ${{t}_{2}}$
So,
\[\Rightarrow {{t}_{2}}-{{t}_{1}}=\dfrac{2\sqrt{{{u}^{2}}-2gh}}{g}\]
\[\Rightarrow 2=\dfrac{2\sqrt{{{u}^{2}}-2gh}}{g}\]
\[\Rightarrow {{u}^{2}}=2gh+{{g}^{2}}\]
\[\Rightarrow \dfrac{u}{g}=\sqrt{1+\dfrac{2h}{g}}\]
\[\Rightarrow \dfrac{u}{g}=\sqrt{1+\dfrac{2h}{g}}\]
Also,
\[\Rightarrow T=\dfrac{2u}{g}=2\sqrt{1+\dfrac{2h}{g}}\]
Where, T is total time of motion
\[\Rightarrow T=2\sqrt{1+\dfrac{2\times 40}{10}}\]
\[\Rightarrow T=2\times 3\]
\[\Rightarrow T=6s\]

Note:
The relation, \[T=\dfrac{2u}{g}\] can easily be derived from the third equation of motion by taking the total displacement equal to zero. This is due to the reason that the point of start is the same as the point of end for the given motion.