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
\[ (\text{Acceleration due to gravity }=10\ \text{ms}^{-2}) \]

• A. \(8\ \text{ms}^{-1}\)
• B. \(40\ \text{ms}^{-1}\)
• C. \(16\ \text{ms}^{-1}\)
• D. \(20\ \text{ms}^{-1}\)

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.

Answer 1

The Correct Option is D

Step 1: Recall the standard projectile trajectory equation.
$y = x\tan\theta - \frac{gx^2}{2u^2\cos^2\theta}$. Comparing with the given equation extracts the projection angle and speed.
Step 2: Find the angle from the coefficient of $x$.
Given $y = \frac{x}{\sqrt{3}} - \frac{x^2}{60}$. Comparing: $\tan\theta = \frac{1}{\sqrt{3}}$, so $\theta = 30^\circ$ and $\cos 30^\circ = \frac{\sqrt{3}}{2}$.
Step 3: Find the range by setting $y = 0$.
\[ 0 = x\!\left(\frac{1}{\sqrt{3}} - \frac{x}{60}\right) \Rightarrow x = 0 \text{ or } x = 20\sqrt{3} \text{ m} \] So $R = 20\sqrt{3}$ m.
Step 4: Apply the range formula to find $u^2$.
With $\sin 60^\circ = \frac{\sqrt{3}}{2}$ and $g = 10$: \[ 20\sqrt{3} = \frac{u^2 \cdot \frac{\sqrt{3}}{2}}{10} \Rightarrow u^2 = 400 \]
Step 5: Compute $u$.
$u = \sqrt{400} = 20$ ms$^{-1}$.
Step 6: State the final answer.
\[ \boxed{u = 20 \text{ ms}^{-1}} \]