Question

In a sports competition, a javelin is thrown at an angle \(45^\circ\), which recorded a range of \(90\text{ m}\). The maximum height reached by the javelin is
\[ \text{(Neglect air resistance and acceleration due to gravity }=10\text{ ms}^{-2}\text{)} \]

• A. \(45\text{ m}\)
• B. \(30\text{ m}\)
• C. \(22.5\text{ m}\)
• D. \(30\sqrt{2}\text{ m}\)

Hint:

For a projectile thrown at \(45^\circ\), \[ R=\frac{u^2}{g} \] which simplifies calculations quickly.

Answer 1

The Correct Option is C

Step 1: Write the range and height formulas.
Range: $R = \frac{u^2 \sin 2\theta}{g}$. Maximum height: $H = \frac{u^2 \sin^2\theta}{2g}$. We form their ratio to avoid computing $u$ separately.
Step 2: Compute the ratio $H/R$.
\[ \frac{H}{R} = \frac{\sin^2\theta}{2\sin 2\theta} \]
Step 3: Substitute $\theta = 45^\circ$.
$\sin^2 45^\circ = \frac{1}{2}$, $\sin 90^\circ = 1$: \[ \frac{H}{R} = \frac{1/2}{2} = \frac{1}{4} \]
Step 4: Solve for $H$ using $R = 90$ m.
\[ H = \frac{R}{4} = \frac{90}{4} = 22.5 \text{ m} \]
Step 5: Verify directly.
From $R = u^2/g$: $u^2 = 900$. Then $H = \frac{900 \times 1/2}{20} = 22.5$ m. Both methods agree.
Step 6: State the final answer.
\[ \boxed{H = 22.5 \text{ m}} \]