A particle is projected vertically upward from the ground. It is observed that it covers equal distance in the 3rd and 8th second. The speed of the particle is 30 m/s at that time.

Steps to Solve

1. Understanding the distances covered in the 3rd and 8th seconds

We can use the formula for the distance covered in the $n^\text{th}$ second given by: $$ d_n = u + \frac{1}{2} g(2n - 1) $$ where (d_n) is the distance covered in the nth second, (u) is the initial velocity, and (g) is the acceleration due to gravity, approximately (9.8 , \text{m/s}^2).

2. Calculating distances for the 3rd and 8th seconds

We set up the equations for the distances:

• For the 3rd second: $$ d_3 = u + \frac{1}{2} g(5) = u + \frac{5g}{2} $$
• For the 8th second: $$ d_8 = u + \frac{1}{2} g(15) = u + \frac{15g}{2} $$

3. Setting the distances equal

Since it is given that the distances covered in the 3rd and 8th seconds are equal, we have: $$ u + \frac{5g}{2} = u + \frac{15g}{2} $$

4. Simplifying the equation

By simplifying the equation, we can cancel (u) from both sides: $$ \frac{5g}{2} = \frac{15g}{2} $$ This results in: $$ 5g = 15g $$ This is incorrect, so we switch perspectives a bit.

Note that: $$ d_3 = d_8 \Rightarrow 5g = 15g \implies g(15 - 5) = 0 $$

5. Using the values given in the problem

Now we also need to verify the given speed condition: If the speed is (30 , \text{m/s}) at the time of interest, $$ v = u - gt $$ By substituting (t) and using (u) to express it using distance formulas again, we can eventually find (u).

6. Using the conditions to solve for (u)

From earlier conclusions, Using (u = g(t) + 30):

Now rearrange and plug the appropriate known values to find feasible (u).