Question

The vertical displacement (y in metre) of a projectile in terms of its horizontal displacement (x in metre) is given by $y=(\sqrt{3}x - 0.2x^2)$. The time of flight of the projectile is (Acceleration due to gravity = 10 ms$^{-2}$)

• A. $5\sqrt{3}$s
• B. $\sqrt{3}$s
• C. 0.2s
• D. $0.2\sqrt{3}$s

Hint:

The equation of a projectile's trajectory, $y = (\tan\theta)x - \frac{g}{2u^2\cos^2\theta}x^2$, is a powerful tool. By comparing a given equation to this standard form, you can quickly extract the launch angle $\theta$ and initial speed $u$.

Answer 1

The Correct Option is B

Step 1: Compare the given trajectory equation with the standard form.
The standard equation of a projectile's trajectory is: \[ y = (\tan\theta)x - \left(\frac{g}{2u^2\cos^2\theta}\right)x^2. \] where $\theta$ is the launch angle and $u$ is the initial speed. The given equation is $y = \sqrt{3}x - 0.2x^2 = \sqrt{3}x - \frac{1}{5}x^2$.

Step 2: Find the launch angle and initial speed.
By comparing the coefficients: $\tan\theta = \sqrt{3} \implies \theta = 60^\circ$. $\frac{g}{2u^2\cos^2\theta} = \frac{1}{5}$. Substitute $g=10$ and $\theta=60^\circ$ (so $\cos\theta=1/2$): \[ \frac{10}{2u^2(1/2)^2} = \frac{1}{5} \implies \frac{10}{2u^2(1/4)} = \frac{1}{5} \implies \frac{10}{u^2/2} = \frac{1}{5}. \] \[ \frac{20}{u^2} = \frac{1}{5} \implies u^2 = 100 \implies u = 10 \text{ m/s}. \]

Step 3: Calculate the time of flight.
The formula for the time of flight, $T$, is: \[ T = \frac{2u\sin\theta}{g}. \] Substitute the values we found: $u=10, \theta=60^\circ, g=10$. \[ T = \frac{2(10)\sin(60^\circ)}{10} = 2\sin(60^\circ). \] \[ T = 2\left(\frac{\sqrt{3}}{2}\right) = \sqrt{3} \text{ s}. \] \[ \boxed{T = \sqrt{3} \text{ s}}. \]