Question

What is the ratio of maximum height attained to the height attained at \( t = 1 \) s for a projectile of initial velocity \( u \) projected at an angle \( 30^\circ \) with horizontal?

Hint:

When dealing with projectile motion, break down the motion into horizontal and vertical components, and use the equations for height and range accordingly.

Answer 1

Step 1: Maximum height formula.
For a projectile launched at an angle \( \theta \) with initial velocity \( u \), the maximum height \( H_{\text{max}} \) is given by: \[ H_{\text{max}} = \frac{u^2 \sin^2 \theta}{2g} \] For \( \theta = 30^\circ \), we have \( \sin 30^\circ = \frac{1}{2} \), so: \[ H_{\text{max}} = \frac{u^2 \left( \frac{1}{2} \right)^2}{2g} = \frac{u^2}{8g} \]
Step 2: Height at \( t = 1 \) second.
The height \( h(t) \) at any time \( t \) is given by: \[ h(t) = u \sin \theta \cdot t - \frac{1}{2} g t^2 \] At \( t = 1 \) second, for \( \theta = 30^\circ \), we have \( \sin 30^\circ = \frac{1}{2} \), so: \[ h(1) = u \cdot \frac{1}{2} \cdot 1 - \frac{1}{2} g \cdot 1^2 = \frac{u}{2} - \frac{g}{2} \]
Step 3: Ratio of heights.
The required ratio of the maximum height to the height at \( t = 1 \) second is: \[ \text{Ratio} = \frac{H_{\text{max}}}{h(1)} = \frac{\frac{u^2}{8g}}{\frac{u}{2} - \frac{g}{2}} \] Simplifying: \[ \text{Ratio} = \frac{\frac{u^2}{8g}}{\frac{u - g}{2}} = \frac{u^2}{8g} \cdot \frac{2}{u - g} \] \[ \text{Ratio} = \frac{u^2}{4g(u - g)} \]