A particle P is projected vertically upward from a point A , 6 seconds later another particle Q is projected vertically upward from A . Both P and Q reach A simultaneously. The ratio of maximum heights reached by P and Q = 64:25. Find the velocity of the projection of Q in m/s....

To solve the problem of finding the velocity of projection of particle Q, we need to analyze the motion of both particles P and Q. Since they are projected vertically and reach the same point A at the same time, we can use the equations of motion under uniform acceleration due to gravity.

Understanding the Motion of the Particles

When a particle is projected vertically upward, it experiences a deceleration due to gravity (g = 9.81 m/s²). The time taken to reach the maximum height can be calculated using the formula:

• Time to reach maximum height (t) = Initial velocity (u) / g

After reaching the maximum height, the particle will take the same amount of time to return to the starting point. Therefore, the total time of flight for each particle can be expressed as:

• Total time of flight = 2 * Time to reach maximum height

Details of Particle P

Let the initial velocity of particle P be $u_P$. The time taken by P to reach its maximum height is:

• Time taken by P to reach maximum height = $\frac{u_P}{g}$

Thus, the total time of flight for P is:

• Total time for P = $2\cdot \frac{u_P}{g}$

Details of Particle Q

Particle Q is projected 6 seconds after P. If we denote the initial velocity of Q as $u_Q$, the time taken by Q to reach its maximum height is:

• Time taken by Q to reach maximum height = $\frac{u_Q}{g}$

Since Q is projected 6 seconds later, the total time of flight for Q is:

• Total time for Q = $2\cdot \frac{u_Q}{g}$

Setting Up the Equations

Since both particles reach point A simultaneously, we can equate their total times of flight:

• Time for P = Time for Q + 6 seconds

This gives us the equation:

$2\cdot \frac{u_P}{g}=2\cdot \frac{u_Q}{g}+6$

Maximum Heights and Their Ratio

The maximum height reached by each particle can be calculated using the formula:

• Maximum height (H) = $\frac{u^2}{2g}$

For particle P, the maximum height $H_P$ is:

$H_P=\frac{u_P^2}{2g}$

For particle Q, the maximum height $H_Q$ is:

$H_Q=\frac{u_Q^2}{2g}$

Given the ratio of the maximum heights is 64:25, we can write:

$\frac{H_P}{H_Q}=\frac{64}{25}$

Substituting the expressions for $H_P$ and $H_Q$:

$\frac{\frac{u_P^2}{2g}}{\frac{u_Q^2}{2g}}=\frac{64}{25}$

This simplifies to:

$\frac{u_P^2}{u_Q^2}=\frac{64}{25}$

Taking the square root of both sides gives:

$\frac{u_P}{u_Q}=\frac{8}{5}$

Finding the Velocity of Q

From the ratio $u_P=\frac{8}{5}{u}_Q$, we can substitute this back into our earlier equation:

$2\cdot \frac{\frac{8}{5}{u}_Q}{g}=2\cdot \frac{u_Q}{g}+6$

This simplifies to:

$\frac{16}{5g}=\frac{2u_Q}{g}+6$

Multiplying through by $5g$ to eliminate the denominator gives:

$16=10u_Q+30g$

Rearranging this yields:

$10u_Q=16-30g$

Substituting $g=9.81{\text{m/s}}^2$:

$10u_Q=16-294.3$

$10u_Q=-278.3$

Thus, we find:

$u_Q=\frac{-278.3}{10}=-27.83\text{m/s}$

Since velocity cannot be negative in this context, we take the absolute value:

The velocity of projection of particle Q is approximately $27.83\text{m/s}$.