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
\[
\text{(Neglect air resistance and acceleration due to gravity }=10\text{ ms}^{-2}\text{)}
\]
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
\[ \text{(Neglect air resistance and acceleration due to gravity }=10\text{ ms}^{-2}\text{)} \]
\[ \text{(Neglect air resistance and acceleration due to gravity }=10\text{ ms}^{-2}\text{)} \]
Show Hint
For a projectile thrown at \(45^\circ\),
\[
R=\frac{u^2}{g}
\]
which simplifies calculations quickly.
Updated On: Jun 25, 2026
- \(45\text{ m}\)
- \(30\text{ m}\)
- \(22.5\text{ m}\)
- \(30\sqrt{2}\text{ m}\)
Show Solution
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The Correct Option is C
Solution and Explanation
Step 1: Use the formula for horizontal range.
For projectile motion, \[ R=\frac{u^2\sin 2\theta}{g} \] Given: \[ R=90\text{ m},\quad \theta=45^\circ,\quad g=10\text{ ms}^{-2} \] Since \[ \sin 2\theta=\sin 90^\circ=1, \] we get \[ 90=\frac{u^2}{10} \] Therefore, \[ u^2=900 \] \[ u=30\text{ ms}^{-1} \]
Step 2: Use the formula for maximum height.
Maximum height is \[ H=\frac{u^2\sin^2\theta}{2g} \] Substitute the values: \[ H=\frac{900\sin^2 45^\circ}{2(10)} \] Since \[ \sin 45^\circ=\frac{1}{\sqrt{2}}, \] we have \[ \sin^2 45^\circ=\frac12 \] Thus, \[ H=\frac{900\times \frac12}{20} \] \[ H=\frac{450}{20} \] \[ H=22.5\text{ m} \]
Step 3: Final conclusion.
Hence, the maximum height reached by the javelin is \[ \boxed{22.5\text{ m}} \]
For projectile motion, \[ R=\frac{u^2\sin 2\theta}{g} \] Given: \[ R=90\text{ m},\quad \theta=45^\circ,\quad g=10\text{ ms}^{-2} \] Since \[ \sin 2\theta=\sin 90^\circ=1, \] we get \[ 90=\frac{u^2}{10} \] Therefore, \[ u^2=900 \] \[ u=30\text{ ms}^{-1} \]
Step 2: Use the formula for maximum height.
Maximum height is \[ H=\frac{u^2\sin^2\theta}{2g} \] Substitute the values: \[ H=\frac{900\sin^2 45^\circ}{2(10)} \] Since \[ \sin 45^\circ=\frac{1}{\sqrt{2}}, \] we have \[ \sin^2 45^\circ=\frac12 \] Thus, \[ H=\frac{900\times \frac12}{20} \] \[ H=\frac{450}{20} \] \[ H=22.5\text{ m} \]
Step 3: Final conclusion.
Hence, the maximum height reached by the javelin is \[ \boxed{22.5\text{ m}} \]
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