The trajectory of a projectile in a vertical plane is y = alpha x - beta x^2, where alpha and beta a

Q: Why do we compare the given equation with the standard trajectory equation?

The standard equation $y = x \tan\theta - \frac{gx^2}{2u^2 \cos^2\theta}$ is the universal mathematical model for any projectile in a vacuum. By comparing a specific equation to it, we can extract the physical parameters like launch angle and initial velocity.

Q: How did we get $\alpha = \tan\theta$?

When two polynomials are equal for all values of $x$, their corresponding coefficients must be equal. By matching the coefficient of the linear $x$ term in both equations, we get $\alpha = \tan\theta$.

Q: What is the physical significance of the $\beta$ term?

The $\beta$ term, which is $\frac{g}{2u^2 \cos^2\theta}$, dictates the 'width' or 'spread' of the parabola. It is inversely proportional to the square of the horizontal component of velocity.

Q: Why did we divide $\beta$ by $\alpha^2$?

It's a mathematical strategy to eliminate the $\cos^2\theta$ term and introduce a $\sin^2\theta$ term, which is required to form the expression for maximum height $H$.

Q: Can we find the horizontal range from this given equation?

Yes! The range $R$ is the non-zero value of $x$ when $y = 0$. Setting $\alpha x - \beta x^2 = 0$ gives $x(\alpha - \beta x) = 0$, so the range $R = \frac{\alpha}{\beta}$.

Question

The trajectory of a projectile in a vertical plane is $y = \alpha x - \beta x^2$, where $\alpha$ and $\beta$ are constants and $x$ and $y$ are respectively the horizontal and vertical distances of the projectile from the point of projection. The angle of projection $	heta$ and the maximum height attained $H$ are respectively given by

Options:

(A) $	an^{-1} \alpha, \frac{\alpha^2}{4\beta}$

(B) $	an^{-1} \beta, \frac{\alpha^2}{2\beta}$

(C) $	an^{-1} \alpha, \frac{4\alpha^2}{\beta}$

(D) $	an^{-1} \left(\frac{\beta}{\alpha}ight), \frac{\alpha^2}{\beta}$

Sigma Percentile · Accepted Answer

### Decoding the Trajectory Equation

Imagine a projectile launched into the air. It traces a beautiful parabolic path. In physics, we can describe this entire journey using a single mathematical equation known as the **equation of trajectory**.

The problem provides us with a specific trajectory equation:

$$y = \alpha x - \beta x^2$$

To unlock the secrets hidden within this equation, we must compare it to the universal standard equation of trajectory for any projectile launched from the origin:

$$y = x 	an	heta - \frac{gx^2}{2u^2 \cos^2	heta}$$

Here, $	heta$ is the angle of projection, $u$ is the initial velocity, and $g$ is the acceleration due to gravity.

### Extracting the Angle of Projection

By placing our two equations side-by-side, we can perform a direct comparison of coefficients. Let's look at the linear term, $x$.

In our given equation, the coefficient is $\alpha$. In the standard equation, it is $	an	heta$. Therefore, we can immediately state:

$$\alpha = 	an	heta$$

Solving for the angle of projection, $	heta$, we get our first crucial result:

$$	heta = 	an^{-1}\alpha$$

### Unveiling the Maximum Height

Next, we compare the coefficients of the quadratic term, $x^2$. This gives us the expression for $\beta$:

$$\beta = \frac{g}{2u^2 \cos^2	heta}$$

Our goal is to find the maximum height, $H$. We know the standard kinematic formula for maximum height is:

$$H = \frac{u^2 \sin^2	heta}{2g}$$

We need to somehow transform our expression for $\beta$ into the expression for $H$. Notice that $\beta$ contains a $\cos^2	heta$ in the denominator, while $H$ requires a $\sin^2	heta$ in the numerator.

Here is a brilliant mathematical trick: Let's divide $\beta$ by $\alpha^2$. Since $\alpha = 	an	heta$, we know that $\alpha^2 = \frac{\sin^2	heta}{\cos^2	heta}$.

$$\frac{\beta}{\alpha^2} = \frac{\frac{g}{2u^2 \cos^2	heta}}{\frac{\sin^2	heta}{\cos^2	heta}}$$

When we perform this division, the $\cos^2	heta$ terms beautifully cancel out!

$$\frac{\beta}{\alpha^2} = \frac{g}{2u^2 \sin^2	heta}$$

Now, look closely at the right side of this equation. It is remarkably similar to the inverse of our maximum height formula. Let's rearrange it to make the connection explicit:

$$\frac{\beta}{\alpha^2} = \frac{1}{4} \left( \frac{2g}{u^2 \sin^2	heta} ight)$$

Since $\frac{1}{H} = \frac{2g}{u^2 \sin^2	heta}$, we can substitute this directly:

$$\frac{\beta}{\alpha^2} = \frac{1}{4} 	imes \frac{1}{H}$$

Finally, solving for $H$, we arrive at our second result:

$$H = \frac{\alpha^2}{4\beta}$$

By simply comparing coefficients and applying a clever algebraic manipulation, we have successfully extracted both the angle of projection and the maximum height from the given trajectory equation.

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Visualized Solution

$Visualizing the Trajectory$

$Standard Equation of Trajectory$

$Comparing Equations$

$Finding Angle of Projection (θ)$

$Extracting β$

$Formula for Maximum Height (H)$

$Relating α, β and H$

$Solving for H$

$Final Conclusion$

The Sigma Insight: Projectile Motion

Decoding the Trajectory Equation

Extracting the Angle of Projection

Unveiling the Maximum Height

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Trajectory

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Standard Equation of Trajectory

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Comparing Equations

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Angle of Projection (θ)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Extracting β

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Formula for Maximum Height (H)

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Relating α,β and H

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for H

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Final Conclusion