MHB Max Horizontal Displacement of Projectile and its Velocity at T

[Theia]

Let $$x = v_0\cos \alpha _0 t$$ and $$y = y_0 + v_0 \sin \alpha _0 t - \tfrac{1}{2} gt^2$$, where

• $$v_0$$ is speed at time $$t = 0$$,
• $$\alpha _0$$ is the angle between positive $$x$$-axis and initial velocity vector ($$\alpha _0 \in (0, \pi/2)$$),
• $$t$$ time in seconds,
• $$y_0 >0$$ the $$y$$ coordinate at time $$t=0$$,
• $$g$$ acceleration due the gravity.

Let $$T$$ be the time when the projectile hits positive $$x$$-axis (i.e. the ground). Find the maximum horizontal displacement of the projectile and show that angle between initial velocity vector and velocity vector at time $$T$$ is $$\pi/2$$.

 

[skeeter]

$x_{max} = \dfrac{v_0^2}{g\tan{\alpha_0}}$

(bonus info that can be determined in working this problem) ... the value of $\alpha_0$ that yields a maximum horizontal range is $\alpha_0 = \arctan\left(\dfrac{v_0}{\sqrt{v_0^2 + 2gy_0}}\right)$

Spoiler

let $a = \alpha_0$, initial launch angle (too lazy to type out the Latex)

$b$ = final impact angle at time $T$$T=\dfrac{x}{v_0\cos{a}}$ substitute this expression for $T$ into the equation for $y$. Note $y(T)=0$

$0=y_0 + x\tan{a} - \dfrac{g}{2v_0^2} \cdot x^2\sec^2{a}$

derivative w/respect to $a$ ...

$0 = x\sec^2{a} + \tan{a} \cdot \dfrac{dx}{da} - \dfrac{g}{v_0^2} \left(x^2 \sec^2{a}\tan{a}+x\sec^2{a} \cdot \dfrac{dx}{da} \right)$

when $x$ is a maximum, $\dfrac{dx}{da} = 0$ ...

$0 = x\sec^2{a} - \dfrac{g}{v_0^2} \left(x^2 \sec^2{a}\tan{a}\right)$

$0 = x\sec^2{a}\bigg[1 - \dfrac{g}{v_0^2} \left(x \tan{a}\right) \bigg]$

$x\sec^2{a} \ne 0 \implies x = \dfrac{v_0^2}{g\tan{a}} \implies T = \dfrac{v_0}{g\sin{a}}$

$v_{fx} = v_0\cos{a}$

$v_{fy} = v_0\sin{a} - gT = -v_0 \cdot \dfrac{\cos^2{a}}{\sin{a}}$

$\tan{b} = \dfrac{v_{fy}}{v_{fx}} = -\cot{a}$

$\tan{b} \cdot \tan{a} = -\cot{a} \cdot \tan{a} = -1 \implies a \perp b$

 

[Theia]

Thank you for your solution! ^^