A body is dropped from a certain height h and the 2^nd body is thrown downwards with velocity 5m/s simultaneously. What would be the difference in height after 3 second.
(a) 15m
(b) 4m
(c) 45m
(d) 12m

Hint: First body was launched with 0 velocity so it would reach some finite distance which would be less, if the same body is launched from the same initial position with some non-zero velocity and travel for the same time as taken by the first body. Since, both bodies go through the same acceleration due to gravity.

Formula Used:
1.Second law of motion given by $s = ut + \dfrac{1}{2}a{t^2}$ …… (1)
Where,
s = total displacement
a=acceleration of the moving body
u= initial velocity of the body

Complete step by step answer:
Given,
Initial velocity for particle 1:${u_1}$
Downward acceleration of the particle 1: ${a_1} = g$ , where, g is acceleration due to gravity.
Initial velocity for particle 2:${u_2} = 5m/s$
Downward acceleration of the particle 2: ${a_2} = g$
Total time of travel for both the particles= $t = 3$sec (Since, both particles were thrown simultaneously)
Step 1:
Using equation (1) we can find total displacement of each particle for time t=3 second.
For particle 1:
Substitute values given, in equation 1 we get-
${s_1} = {u_1}t + \dfrac{1}{2}{a_1}{t^2}$
$ \Rightarrow {s_1} = 0t + \dfrac{1}{2}g{t^2}$ $ \Rightarrow {s_1} = \dfrac{1}{2} \times 10 \times {3^2}$
$ \Rightarrow {s_1} = 135m$ …… (2)
Step 2:
Similarly, we can find total displacement for particle 2 as well.
For particle 2:
Substituting given values for particle 2 in equation (1) we get-
$ \Rightarrow {s_2} = {u_2}t + \dfrac{1}{2}{a_2}{t^2}$
$ \Rightarrow {s_2} = {u_2}t + \dfrac{1}{2}g{t^2}$ $ \Rightarrow {s_2} = 5 \times 3 + \dfrac{1}{2} \times 10 \times {3^2} \Rightarrow {s_2} - {s_1} = 15m$
$ \Rightarrow {s_2} = 150m$ …… (3)
Step 3:
Subtracting equation (2) from equation (2) we get-
$ \Rightarrow {s_2} - {s_1} = 150m - 135m$
So, gap between two particles=

Final Answer
Hence, option (a) 15m is correct


Note: This can be done in shortcut way as well:
Method 2:
For this just subtract the total distance relation for particles directly before supplying values into the equation.
$ \Rightarrow {s_2} - {s_1} = {u_2}t + \dfrac{1}{2}{a_2}{t^2} - {u_1}t + \dfrac{1}{2}{a_1}{t^2}$
We get, $ \Rightarrow {s_2} - {s_1} = ({u_2} - {u_1})t$
Substitute, values of initial velocities we get, $ \Rightarrow {s_2} - {s_1} = 15m$