Question:
The equation for the trajectory of a projectile is
\[
y=\left(\frac{x}{\sqrt{3}}-\frac{x^2}{60}\right)\text{ m}
\]
The velocity of projection of the projectile is
\[
(\text{Acceleration due to gravity }=10\ \text{ms}^{-2})
\]
The equation for the trajectory of a projectile is
\[
y=\left(\frac{x}{\sqrt{3}}-\frac{x^2}{60}\right)\text{ m}
\]
The velocity of projection of the projectile is
\[ (\text{Acceleration due to gravity }=10\ \text{ms}^{-2}) \]
\[ (\text{Acceleration due to gravity }=10\ \text{ms}^{-2}) \]
Show Hint
The standard trajectory equation of a projectile is
\[
y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta}
\]
Compare coefficients carefully to determine the angle and velocity of projection.
Updated On: Jun 25, 2026
- \(8\ \text{ms}^{-1}\)
- \(40\ \text{ms}^{-1}\)
- \(16\ \text{ms}^{-1}\)
- \(20\ \text{ms}^{-1}\)
Show Solution
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The Correct Option is D
Solution and Explanation
Step 1: Write the standard equation of trajectory.
The trajectory of a projectile is given by \[ y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta} \] where \[ u=\text{velocity of projection} \] and \[ \theta=\text{angle of projection} \]
Step 2: Compare the given equation with the standard form.
Given equation: \[ y=\frac{x}{\sqrt{3}}-\frac{x^2}{60} \] Comparing the coefficient of \(x\), \[ \tan\theta=\frac{1}{\sqrt{3}} \] Hence, \[ \theta=30^\circ \] Now, \[ \cos30^\circ=\frac{\sqrt{3}}{2} \] Therefore, \[ \cos^2 30^\circ=\frac{3}{4} \]
Step 3: Compare the coefficient of \(x^2\).
From the trajectory equation, \[ \frac{g}{2u^2\cos^2\theta}=\frac{1}{60} \] Substituting \[ g=10 \] and \[ \cos^2\theta=\frac{3}{4}, \] we get \[ \frac{10}{2u^2\left(\frac{3}{4}\right)}=\frac{1}{60} \] \[ \frac{10}{\frac{3u^2}{2}}=\frac{1}{60} \] \[ \frac{20}{3u^2}=\frac{1}{60} \] Cross multiplying, \[ 20\times 60=3u^2 \] \[ 1200=3u^2 \] \[ u^2=400 \] \[ u=20\ \text{ms}^{-1} \]
Step 4: Final conclusion.
Hence, the velocity of projection is \[ \boxed{20\ \text{ms}^{-1}} \]
The trajectory of a projectile is given by \[ y=x\tan\theta-\frac{gx^2}{2u^2\cos^2\theta} \] where \[ u=\text{velocity of projection} \] and \[ \theta=\text{angle of projection} \]
Step 2: Compare the given equation with the standard form.
Given equation: \[ y=\frac{x}{\sqrt{3}}-\frac{x^2}{60} \] Comparing the coefficient of \(x\), \[ \tan\theta=\frac{1}{\sqrt{3}} \] Hence, \[ \theta=30^\circ \] Now, \[ \cos30^\circ=\frac{\sqrt{3}}{2} \] Therefore, \[ \cos^2 30^\circ=\frac{3}{4} \]
Step 3: Compare the coefficient of \(x^2\).
From the trajectory equation, \[ \frac{g}{2u^2\cos^2\theta}=\frac{1}{60} \] Substituting \[ g=10 \] and \[ \cos^2\theta=\frac{3}{4}, \] we get \[ \frac{10}{2u^2\left(\frac{3}{4}\right)}=\frac{1}{60} \] \[ \frac{10}{\frac{3u^2}{2}}=\frac{1}{60} \] \[ \frac{20}{3u^2}=\frac{1}{60} \] Cross multiplying, \[ 20\times 60=3u^2 \] \[ 1200=3u^2 \] \[ u^2=400 \] \[ u=20\ \text{ms}^{-1} \]
Step 4: Final conclusion.
Hence, the velocity of projection is \[ \boxed{20\ \text{ms}^{-1}} \]
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