A particle is projected at an angle of 30° to the Horizontal with a speed of 250m/s. Calculate the time of flight, speed of the particle at its maximum height. (g = 10m/s)

$$T = \frac{2 \times 250 \, \text{m/s} \times \sin(30^\circ)}{10 \, \text{m/s}^2}$$

We know that $\sin(30^\circ) = 0.5$.

$$T = \frac{2 \times 250 \times 0.5}{10} \, \text{s}$$ $$T = \frac{250}{10} \, \text{s}$$ $$T = 25 \, \text{s}$$

The time of flight is 25 s.

$$\boxed{T = 25 \, \text{s}}$$

The horizontal component of the velocity remains constant throughout the projectile motion (neglecting air resistance).
The horizontal component of the initial velocity ($u_x$) is given by:

$$u_x = u \cos\theta$$

$$u_x = 250 \, \text{m/s} \times \cos(30^\circ)$$

We know that $\cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866$.

$$u_x = 250 \times \frac{\sqrt{3}}{2} \, \text{m/s}$$ $$u_x = 125\sqrt{3} \, \text{m/s}$$ $$u_x \approx 125 \times 1.732 \, \text{m/s}$$ $$u_x \approx 216.5 \, \text{m/s}$$

The speed of the particle at its maximum height is approximately 216.5 m/s.

$$\boxed{u_x = 125\sqrt{3} \, \text{m/s} \approx 216.5 \, \text{m/s}}$$