A particle is projected with a velocity of 196m/s at an angle of 30 to the horizontal. Calculate the; a): greatest height b): time of flight

Step 1: Identify the given values and formulas.
Given:
Initial velocity, $u = 196 \text{ m/s}$
Angle of projection, $\theta = 30^\circ$
Acceleration due to gravity, $g = 9.8 \text{ m/s}^2$

a) Calculate the greatest height.
The formula for the greatest height ($H$) is:

$$H = \frac{u^2 \sin^2 \theta}{2g}$$

Substitute the given values:

$$H = \frac{(196 \text{ m/s})^2 (\sin 30^\circ)^2}{2 \times 9.8 \text{ m/s}^2}$$ $$H = \frac{38416 \text{ m}^2/\text{s}^2 \times (0.5)^2}{19.6 \text{ m/s}^2}$$ $$H = \frac{38416 \text{ m}^2/\text{s}^2 \times 0.25}{19.6 \text{ m/s}^2}$$ $$H = \frac{9604 \text{ m}^2/\text{s}^2}{19.6 \text{ m/s}^2}$$ $$H = 490 \text{ m}$$

The greatest height is $\boxed{\text{490 m}}$.

b) Calculate the time of flight.
The formula for the time of flight ($T$) is:

$$T = \frac{2u \sin \theta}{g}$$

Substitute the given values:

$$T = \frac{2 \times 196 \text{ m/s} \times \sin 30^\circ}{9.8 \text{ m/s}^2}$$ $$T = \frac{2 \times 196 \text{ m/s} \times 0.5}{9.8 \text{ m/s}^2}$$ $$T = \frac{196 \text{ m/s}}{9.8 \text{ m/s}^2}$$ $$T = 20 \text{ s}$$

The time of flight is $\boxed{\text{20 s}}$.