Question

A ball is thrown at speed $v$ from zero height on level ground. Let $\theta_0$ be the angle at which the ball should be thrown so that the length of the trajectory is maximum. If $\theta_0$ satisfies

$ln \left| \frac{\alpha + sin \theta_0}{\beta cos \theta_0} \right| = \frac{\gamma}{sin \theta_0}$

then find the value of $\alpha + \beta + \gamma^2$.

$\left\{ \text{you can use } \int \sqrt{1 + x^2} dx = \frac{x}{2} \sqrt{1 + x^2} + \frac{1}{2} \{n\} |x + \sqrt{1 + x^2}| + C \right\}$

Answer 1

3

Answer 2

To find the angle $\theta_0$ for maximum trajectory length, we first need to derive the expression for the trajectory length $L$.

1. Parametric Equations of Trajectory:

For a projectile launched with initial speed $v$ at an angle $\theta$ from the horizontal, the position coordinates at time $t$ are:

$x(t) = (v \cos \theta) t$

$y(t) = (v \sin \theta) t - \frac{1}{2} g t^2$

The horizontal and vertical components of velocity are:

$\frac{dx}{dt} = v \cos \theta$

$\frac{dy}{dt} = v \sin \theta - gt$

The ball lands when $y(t) = 0$ (excluding $t=0$).

$(v \sin \theta) t - \frac{1}{2} g t^2 = 0$

$t \left( v \sin \theta - \frac{1}{2} g t \right) = 0$

The time of flight $T = \frac{2v \sin \theta}{g}$.

2. Trajectory Length Formula:

The length of the trajectory $L$ is given by the arc length integral:

$L = \int_0^T \sqrt{\left(\frac{dx}{dt}\right)^2 + \left(\frac{dy}{dt}\right)^2} dt$

$L = \int_0^T \sqrt{(v \cos \theta)^2 + (v \sin \theta - gt)^2} dt$

$L = \int_0^T \sqrt{v^2 \cos^2 \theta + v^2 \sin^2 \theta - 2vg t \sin \theta + g^2 t^2} dt$

$L = \int_0^T \sqrt{v^2 - 2vg t \sin \theta + g^2 t^2} dt$

To simplify the integral, let's make a substitution. Notice that the term inside the square root can be written as $(v \cos \theta)^2 + (gt - v \sin \theta)^2$.

Let $X = \frac{gt - v \sin \theta}{v \cos \theta}$. Then $dX = \frac{g}{v \cos \theta} dt$, so $dt = \frac{v \cos \theta}{g} dX$.

Now, change the limits of integration for $X$:

When $t=0$, $X_1 = \frac{0 - v \sin \theta}{v \cos \theta} = -\tan \theta$.

When $t=T = \frac{2v \sin \theta}{g}$, $X_2 = \frac{g \left(\frac{2v \sin \theta}{g}\right) - v \sin \theta}{v \cos \theta} = \frac{2v \sin \theta - v \sin \theta}{v \cos \theta} = \frac{v \sin \theta}{v \cos \theta} = \tan \theta$.

Substituting these into the integral for $L$:

$L = \int_{-\tan \theta}^{\tan \theta} \sqrt{(v \cos \theta)^2 + (v \cos \theta \cdot X)^2} \left(\frac{v \cos \theta}{g}\right) dX$

$L = \int_{-\tan \theta}^{\tan \theta} v \cos \theta \sqrt{1 + X^2} \left(\frac{v \cos \theta}{g}\right) dX$

$L = \frac{v^2 \cos^2 \theta}{g} \int_{-\tan \theta}^{\tan \theta} \sqrt{1 + X^2} dX$

Since $\sqrt{1+X^2}$ is an even function, we can write:

$L = \frac{2v^2 \cos^2 \theta}{g} \int_0^{\tan \theta} \sqrt{1 + X^2} dX$

3. Evaluate the Integral:

Using the given formula $\int \sqrt{1 + x^2} dx = \frac{x}{2} \sqrt{1 + x^2} + \frac{1}{2} \ln |x + \sqrt{1 + x^2}| + C$:

$L = \frac{2v^2 \cos^2 \theta}{g} \left[ \frac{X}{2} \sqrt{1 + X^2} + \frac{1}{2} \ln |X + \sqrt{1 + X^2}| \right]_0^{\tan \theta}$

Evaluate at the limits:

At $X = \tan \theta$:

$\frac{\tan \theta}{2} \sqrt{1 + \tan^2 \theta} + \frac{1}{2} \ln |\tan \theta + \sqrt{1 + \tan^2 \theta}|$

$= \frac{\tan \theta}{2} \sec \theta + \frac{1}{2} \ln |\tan \theta + \sec \theta|$

$= \frac{\sin \theta}{2 \cos^2 \theta} + \frac{1}{2} \ln \left| \frac{\sin \theta}{\cos \theta} + \frac{1}{\cos \theta} \right|$

$= \frac{\sin \theta}{2 \cos^2 \theta} + \frac{1}{2} \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right|$

At $X = 0$:

$\frac{0}{2} \sqrt{1 + 0^2} + \frac{1}{2} \ln |0 + \sqrt{1 + 0^2}| = 0 + \frac{1}{2} \ln |1| = 0$.

So, $L = \frac{2v^2 \cos^2 \theta}{g} \left[ \frac{\sin \theta}{2 \cos^2 \theta} + \frac{1}{2} \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| \right]$

$L = \frac{v^2}{g} \left[ \sin \theta + \cos^2 \theta \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| \right]$

4. Maximize Trajectory Length:

To find the angle $\theta_0$ that maximizes $L$, we need to differentiate $L$ with respect to $\theta$ and set it to zero.

Let $f(\theta) = \sin \theta + \cos^2 \theta \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right|$. We need to find $\frac{df}{d\theta} = 0$.

$\frac{d}{d\theta} (\sin \theta) = \cos \theta$

For the second term, use the product rule and chain rule:

$\frac{d}{d\theta} \left( \cos^2 \theta \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| \right)$

$= \left( \frac{d}{d\theta} \cos^2 \theta \right) \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| + \cos^2 \theta \left( \frac{d}{d\theta} \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| \right)$

$= (-2 \cos \theta \sin \theta) \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| + \cos^2 \theta \left( \frac{d}{d\theta} (\ln(1+\sin\theta) - \ln(\cos\theta)) \right)$

$= -2 \cos \theta \sin \theta \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| + \cos^2 \theta \left( \frac{\cos \theta}{1 + \sin \theta} - \frac{-\sin \theta}{\cos \theta} \right)$

$= -2 \cos \theta \sin \theta \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| + \cos^2 \theta \left( \frac{\cos \theta}{1 + \sin \theta} + \frac{\sin \theta}{\cos \theta} \right)$

$= -2 \cos \theta \sin \theta \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| + \cos^2 \theta \left( \frac{\cos^2 \theta + \sin \theta (1 + \sin \theta)}{(1 + \sin \theta) \cos \theta} \right)$

$= -2 \cos \theta \sin \theta \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| + \cos \theta \left( \frac{\cos^2 \theta + \sin \theta + \sin^2 \theta}{1 + \sin \theta} \right)$

$= -2 \cos \theta \sin \theta \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| + \cos \theta \left( \frac{1 + \sin \theta}{1 + \sin \theta} \right)$

$= -2 \cos \theta \sin \theta \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| + \cos \theta$

Combining the derivatives:

$\frac{df}{d\theta} = \cos \theta - 2 \cos \theta \sin \theta \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| + \cos \theta$

$\frac{df}{d\theta} = 2 \cos \theta - 2 \cos \theta \sin \theta \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right|$

$\frac{df}{d\theta} = 2 \cos \theta \left( 1 - \sin \theta \ln \left| \frac{1 + \sin \theta}{\cos \theta} \right| \right)$

Set $\frac{df}{d\theta} = 0$. Since $\theta_0$ is an angle for projectile motion, $0 < \theta_0 < \pi/2$, so $\cos \theta_0 \neq 0$.

Therefore, we must have:

$1 - \sin \theta_0 \ln \left| \frac{1 + \sin \theta_0}{\cos \theta_0} \right| = 0$

$\sin \theta_0 \ln \left| \frac{1 + \sin \theta_0}{\cos \theta_0} \right| = 1$

$\ln \left| \frac{1 + \sin \theta_0}{\cos \theta_0} \right| = \frac{1}{\sin \theta_0}$

5. Determine $\alpha, \beta, \gamma$ and calculate the final value:

The derived equation for $\theta_0$ is:

$\ln \left| \frac{1 + \sin \theta_0}{\cos \theta_0} \right| = \frac{1}{\sin \theta_0}$

Comparing this with the given equation:

$ln \left| \frac{\alpha + sin \theta_0}{\beta cos \theta_0} \right| = \frac{\gamma}{sin \theta_0}$

By direct comparison, we find:

$\alpha = 1$

$\beta = 1$

$\gamma = 1$

Finally, we need to find the value of $\alpha + \beta + \gamma^2$:

$\alpha + \beta + \gamma^2 = 1 + 1 + 1^2 = 1 + 1 + 1 = 3$.

The final answer is $\boxed{3}$.