Question:
A projectile is projected with speed \(20\ \text{m/s}\) at an angle of \(30^\circ\). The range is \((g=10\ \text{m/s}^2)\):
A projectile is projected with speed \(20\ \text{m/s}\) at an angle of \(30^\circ\). The range is \((g=10\ \text{m/s}^2)\):
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For projectile range, use:
\[
R=\frac{u^2\sin 2\theta}{g}
\]
Always remember that the angle becomes \(2\theta\) inside the sine function.
Updated On: Jun 3, 2026
- \(20\ \text{m}\)
- \(30\ \text{m}\)
- \(40\ \text{m}\)
- \(60\ \text{m}\)
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The Correct Option is B
Solution and Explanation
Concept:
For a projectile projected with speed \(u\) at an angle \(\theta\), the horizontal range is given by: \(\displaystyle R=\frac{u^2\sin 2\theta}{g}\) where, \(R=\) range of projectile
\(u=\) initial speed
\(\theta=\) angle of projection
\(g=\) acceleration due to gravity
Step 1: Write the given values.
Initial speed: \(\displaystyle u=20\ \text{m/s}\) Angle of projection: \(\displaystyle \theta=30^\circ\) Acceleration due to gravity: \(\displaystyle g=10\ \text{m/s}^2\)
Step 2: Use the range formula.
\(\displaystyle R=\frac{u^2\sin 2\theta}{g}\) Substitute the values: \(\displaystyle R=\frac{(20)^2\sin(2\times 30^\circ)}{10}\) \(\displaystyle R=\frac{400\sin 60^\circ}{10}\)
Step 3: Use the value of \(\sin 60^\circ\).
\(\displaystyle \sin 60^\circ=\frac{\sqrt{3}}{2}\) So, \(\displaystyle R=\frac{400}{10}\times \frac{\sqrt{3}}{2}\) \(\displaystyle R=40\times \frac{\sqrt{3}}{2}\) \(\displaystyle R=20\sqrt{3}\)
Step 4: Find approximate value.
Since, \(\displaystyle \sqrt{3}\approx 1.732\) Therefore, \(\displaystyle R=20\times 1.732\) \(\displaystyle R=34.64\ \text{m}\)
Step 5: Choose the nearest option.
The nearest option to \(34.64\ \text{m}\) is: \(\displaystyle 30\ \text{m}\) Hence, the correct answer is: \(\displaystyle \boxed{30\ \text{m}}\)
For a projectile projected with speed \(u\) at an angle \(\theta\), the horizontal range is given by: \(\displaystyle R=\frac{u^2\sin 2\theta}{g}\) where, \(R=\) range of projectile
\(u=\) initial speed
\(\theta=\) angle of projection
\(g=\) acceleration due to gravity
Step 1: Write the given values.
Initial speed: \(\displaystyle u=20\ \text{m/s}\) Angle of projection: \(\displaystyle \theta=30^\circ\) Acceleration due to gravity: \(\displaystyle g=10\ \text{m/s}^2\)
Step 2: Use the range formula.
\(\displaystyle R=\frac{u^2\sin 2\theta}{g}\) Substitute the values: \(\displaystyle R=\frac{(20)^2\sin(2\times 30^\circ)}{10}\) \(\displaystyle R=\frac{400\sin 60^\circ}{10}\)
Step 3: Use the value of \(\sin 60^\circ\).
\(\displaystyle \sin 60^\circ=\frac{\sqrt{3}}{2}\) So, \(\displaystyle R=\frac{400}{10}\times \frac{\sqrt{3}}{2}\) \(\displaystyle R=40\times \frac{\sqrt{3}}{2}\) \(\displaystyle R=20\sqrt{3}\)
Step 4: Find approximate value.
Since, \(\displaystyle \sqrt{3}\approx 1.732\) Therefore, \(\displaystyle R=20\times 1.732\) \(\displaystyle R=34.64\ \text{m}\)
Step 5: Choose the nearest option.
The nearest option to \(34.64\ \text{m}\) is: \(\displaystyle 30\ \text{m}\) Hence, the correct answer is: \(\displaystyle \boxed{30\ \text{m}}\)
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