# A particle is projected at a definite angle alpha to the horizontal passes through the points (a,b) and (b,a), referred to horizontal and vertical axes through the point of projection?

## Calculate range and $\tan \alpha$ in terms of a and b?

Aug 9, 2018

The answers are $\tan \alpha = \frac{{b}^{2} + a b + {a}^{2}}{a b}$ and the range is $= \left(a + b\right)$

#### Explanation:

The trajectory of a projectile in the $x - y$ plane is given by the equation

$y = x \tan \theta - \left(\frac{g}{2 {u}^{2} {\cos}^{2} \theta}\right) {x}^{2}$

Here,

The initial velocity is $= u$

The angle is $\theta = \alpha$

The acceleration due to gravity is $= g$

The points are $= \left(a , b\right)$ and $= \left(b , a\right)$

Therefore,

$b = a \tan \alpha - \left(\frac{g}{2 {u}^{2} {\cos}^{2} \alpha}\right) {a}^{2}$.............$\left(1\right)$

$a = b \tan \alpha - \left(\frac{g}{2 {u}^{2} {\cos}^{2} \alpha}\right) {b}^{2}$.............$\left(2\right)$

Multiply equation $\left(1\right)$ by ${b}^{2}$ and equation $\left(2\right)$ by ${a}^{2}$

${b}^{3} = a {b}^{2} \tan \alpha - \left(\frac{g}{2 {u}^{2} {\cos}^{2} \alpha}\right) {a}^{2} {b}^{2}$.............$\left(3\right)$

${a}^{3} = {a}^{2} b \tan \alpha - \left(\frac{g}{2 {u}^{2} {\cos}^{2} \alpha}\right) {b}^{2} {a}^{2}$.............$\left(4\right)$

Then,

$\left(3\right) - \left(4\right)$ is

${b}^{3} - {a}^{3} = a {b}^{2} \tan \alpha - {a}^{2} b \tan \alpha$

$= a b \left(b - a\right) \tan \alpha$

But,

${b}^{3} - {a}^{3} = \left(b - a\right) \left({b}^{2} + a b + {a}^{2}\right)$

Therefore,

$\tan \alpha = \frac{{b}^{3} - {a}^{3}}{a b \left(b - a\right)} = \frac{\cancel{b - a} \left({b}^{2} + a b + {a}^{2}\right)}{a b \cancel{b - a}}$

$= \frac{{b}^{2} + a b + {a}^{2}}{a b}$

The range is when

$a = 0$ or $b = 0$

Therefore,

$0 = a \tan \alpha - \left(\frac{g}{2 {u}^{2} {\cos}^{2} \alpha}\right) {a}^{2}$.............$\left(3\right)$

$0 = b \tan \alpha - \left(\frac{g}{2 {u}^{2} {\cos}^{2} \alpha}\right) {b}^{2}$.............$\left(4\right)$

Then,

$\left(4\right) - \left(3\right)$ is

$\left(b - a\right) \tan \alpha = \left({b}^{2} - {a}^{2}\right) \left(\frac{g}{2 {u}^{2} {\cos}^{2} \alpha}\right)$

${u}^{2} = \frac{g \left(a + b\right)}{2 \tan \alpha {\cos}^{2} \alpha} = \frac{g \left(a + b\right)}{\sin 2 \alpha}$

The range is

$r = {u}^{2} \sin \frac{2 \alpha}{g} = \frac{g \left(a + b\right)}{\sin 2 \alpha} \cdot \sin \frac{2 \alpha}{g} = a + b$

Aug 9, 2018

• $\text{ Range is: } q \quad \frac{{a}^{2} + {b}^{2} + a b}{a + b}$

• $q \quad \tan \alpha = \frac{{a}^{2} + {b}^{2} + a b}{a b}$

#### Explanation:

Projectile motion is parabolic . This is easy to show, by replacing parameter $t$, in the equations for horizontal and vertical displacements, by the fixed values, namely ${u}_{x} , {u}_{y} , \alpha$:

• $q \quad q \quad \left\{\begin{matrix}x \left(t\right) = {u}_{x} t \\ y \left(t\right) = {u}_{y} t - \frac{1}{2} g {t}^{2} \\ \tan \alpha = {u}_{y} / {u}_{x}\end{matrix}\right.$

So for parabola passing through origin:

• $y = x \left(\lambda x + \mu\right) q \quad \text{ Roots } \to q \quad \left\{\begin{matrix}x = 0 \\ x = \boldsymbol{- \frac{\mu}{\lambda}} q \quad \square\end{matrix}\right.$

It follows from the IV's that:

• {(b = a ( lambda a + mu)),(a = b ( lambda b + mu) ):} qquad " or " qquad [(a,1),(b,1)][(lambda),(mu)] = [(b/a),(a/b)]

Noting the restrictions that make the algebra impossible, or the matrix singular (esp $a \ne b$), this leaves:

$\lambda = - \frac{a + b}{a b} q \quad q \quad \mu = \frac{{a}^{2} + {b}^{2} + a b}{a b}$

The range is the second root of $y \left(x\right)$, as stated in $\square$ above:

$\therefore \text{ Range is: } \frac{{a}^{2} + {b}^{2} + a b}{a + b}$

$\tan \alpha$ is the derivative (wrt $x$) of the parabola at the Origin:

• $\text{ie } q \quad q \quad \tan \alpha = {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}_{x = 0} q \quad \equiv \left(\frac{\dot{y}}{\dot{x}}\right) {\setminus}_{t = 0} q \quad = y ' \left(0\right)$

$y ' = 2 \lambda x + \mu$

$y ' \left(0\right) = \mu = \boldsymbol{\frac{{a}^{2} + {b}^{2} + a b}{a b} = \tan \alpha}$