Step 1: Analyze the Given Data:
Initial velocity, \( u = 19.6 \, \text{ms}^{-1} \).
Maximum height, \( H = 9.8 \, \text{m} \).
Acceleration due to gravity, \( g \approx 9.8 \, \text{ms}^{-2} \).
Step 2: Use the Maximum Height Formula to find angle \( \theta \):
The formula for maximum height is:
\[ H = \frac{u^2 \sin^2 \theta}{2g} \]
Substitute the values:
\[ 9.8 = \frac{(19.6)^2 \sin^2 \theta}{2(9.8)} \]
\[ 9.8 = \frac{19.6 \times 19.6 \times \sin^2 \theta}{19.6} \]
\[ 9.8 = 19.6 \sin^2 \theta \]
\[ \sin^2 \theta = \frac{9.8}{19.6} = \frac{1}{2} \]
\[ \sin \theta = \frac{1}{\sqrt{2}} \implies \theta = 45^\circ \]
Step 3: Calculate the Range (R):
The formula for horizontal range is:
\[ R = \frac{u^2 \sin 2\theta}{g} \]
Since \( \theta = 45^\circ \), \( 2\theta = 90^\circ \) and \( \sin 90^\circ = 1 \).
\[ R = \frac{(19.6)^2 (1)}{9.8} \]
\[ R = \frac{19.6 \times 19.6}{9.8} \]
\[ R = 2 \times 19.6 = 39.2 \, \text{m} \]
Alternative Approach:
We know the relation \( R = 4H \cot \theta \).
From Step 2, we found \( \sin^2 \theta = 1/2 \), so \( \theta = 45^\circ \).
\( \cot 45^\circ = 1 \).
\[ R = 4(9.8)(1) = 39.2 \, \text{m} \]