Question

The trajectory of a projectile in a vertical ple y = ax - bx², where a and b are constants and y respectively are horizontal and distances of the projectile from the po projection. The maximum height attained particle and the angle of projection fror horizontal are

• A. $\frac{b^2}{2a}$, tan⁻¹(b)
• B. $\frac{a^2}{b}$, tan⁻¹(2
• C. $\frac{a^2}{4b}$, tan⁻¹(a)
• D. $\frac{2a^2}{b}$, tan⁻¹

Answer 1

Answer: (C)

$\frac{a^2}{4b}$, tan⁻¹(a)

Answer 2

The standard equation of the trajectory of a projectile is given by: $y = x \tan \theta - \frac{g x^2}{2 u^2 \cos^2 \theta}$ where $\theta$ is the angle of projection with the horizontal, $u$ is the initial velocity, and $g$ is the acceleration due to gravity.

The given equation of the trajectory is: $y = ax - bx^2$

Comparing the coefficients of $x$ and $x^2$ from both equations:

1. Coefficient of $x$: $\tan \theta = a$ From this, the angle of projection $\theta$ is $\tan^{-1}(a)$.

2. Coefficient of $x^2$: $\frac{g}{2 u^2 \cos^2 \theta} = b$

Now, we need to find the maximum height attained by the particle. The formula for maximum height (H) in projectile motion is: $H = \frac{u^2 \sin^2 \theta}{2g}$

From the second comparison, we can express $u^2$ in terms of $b$ and $\cos \theta$: $2 u^2 \cos^2 \theta = \frac{g}{b}$ $u^2 = \frac{g}{2b \cos^2 \theta}$

Substitute this expression for $u^2$ into the maximum height formula: $H = \frac{\left( \frac{g}{2b \cos^2 \theta} \right) \sin^2 \theta}{2g}$ $H = \frac{g \sin^2 \theta}{4bg \cos^2 \theta}$ $H = \frac{\sin^2 \theta}{4b \cos^2 \theta}$ $H = \frac{1}{4b} \left( \frac{\sin \theta}{\cos \theta} \right)^2$ $H = \frac{1}{4b} \tan^2 \theta$

Since we found that $\tan \theta = a$, we can substitute this into the equation for H: $H = \frac{1}{4b} (a)^2$ $H = \frac{a^2}{4b}$

Thus, the maximum height attained by the particle is $\frac{a^2}{4b}$ and the angle of projection is $\tan^{-1}(a)$.

Comparing this with the given options, the calculated values match option (3).