Question

A particle is projected with a speed of 20 m/s at an angle of 30° with the horizontal. Find: The maximum height reached by the particle. The time of flight. The horizontal range. (Take g = 10 m/s²)

Answer 1

The maximum height reached by the particle is 5 m. The time of flight is 2 s. The horizontal range is $20\sqrt{3}$ m.

Answer 2

The problem involves projectile motion. We are given the initial speed ($u$), the angle of projection ($\theta$), and the acceleration due to gravity ($g$). We need to find the maximum height ($H$), time of flight ($T$), and horizontal range ($R$).

Given values: Initial speed, $u = 20$ m/s Angle of projection, $\theta = 30^\circ$ Acceleration due to gravity, $g = 10$ m/s$^2$

We use the standard formulas for projectile motion:

1. Maximum Height (H): The formula for maximum height is: $H = \frac{u^2 \sin^2 \theta}{2g}$ Substituting the given values: $\sin 30^\circ = \frac{1}{2}$ $\sin^2 30^\circ = (\frac{1}{2})^2 = \frac{1}{4}$ $H = \frac{(20)^2 \times \frac{1}{4}}{2 \times 10} = \frac{400 \times \frac{1}{4}}{20} = \frac{100}{20} = 5 \text{ m}$

2. Time of Flight (T): The formula for the time of flight is: $T = \frac{2u \sin \theta}{g}$ Substituting the given values: $T = \frac{2 \times 20 \times \frac{1}{2}}{10} = \frac{20}{10} = 2 \text{ s}$

3. Horizontal Range (R): The formula for the horizontal range is: $R = \frac{u^2 \sin(2\theta)}{g}$ Substituting the given values: $\sin(2\theta) = \sin(2 \times 30^\circ) = \sin(60^\circ) = \frac{\sqrt{3}}{2}$ $R = \frac{(20)^2 \times \frac{\sqrt{3}}{2}}{10} = \frac{400 \times \frac{\sqrt{3}}{2}}{10} = \frac{200\sqrt{3}}{10} = 20\sqrt{3} \text{ m}$

Summary of Results:

• The maximum height reached by the particle is 5 m.
• The time of flight is 2 s.
• The horizontal range is $20\sqrt{3}$ m.