Answer

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**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$

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