Question

A projectile is fired up an incline (incline angle $\mathit{\varphi }$) with an initial speed ${\mathit{v}}_{\mathbf{i}}$ at an angle ${\mathit{\theta }}_{\mathbf{i}}$ with respect to the horizontal $\mathbf{(}{\mathit{\theta }}_{\mathbf{i}}\mathbf{>}\mathit{\varphi }\mathbf{)}$as shown in Figure P4.86.

(a) Show that the projectile travels a distance$\mathit{d}$up the incline, where

$\mathit{d}\mathbf{=}\frac{\mathbf{2}{\mathbf{v}}_{\mathbf{i}}^{\mathbf{2}}\mathbf{c}\mathbf{o}\mathbf{s}{\mathbf{\theta }}_{\mathbf{i}}\mathbf{s}\mathbf{i}\mathbf{n}\mathbf{(}{\mathbf{\theta }}_{\mathbf{i}}\mathbf{-}\mathbf{\varphi }\mathbf{)}}{\mathbf{g}\mathbf{c}\mathbf{o}{\mathbf{s}}^{\mathbf{2}}\mathbf{\varphi }}$

(b) For what value of ${\mathit{\theta }}_{\mathbf{i}}$is $\mathit{d}$a maximum, and what is that maximum value?

Short answer

1. It is proven that$d=\frac{2v_i^2\cos {\theta }_i\sin \left({\theta }_i-f\right)}{g{\cos }^{2}f}$
2. is maximum at $=\frac{{\mathrm{v}}_{\mathrm{i}}^2(1-\sin \varphi )}{{\mathrm{gcos}}^2\varphi }$

Step-by-step solution

Given data

(a) Initial velocity = $v_i$

(b) An angle =${\theta }_i$

Assumeand vertical components of the velocity, then

The $x$-component is:

$v_{xi}=v_i\cos {\theta }_i$

The $y$component is:

$v_{yi}=v_i\sin {\theta }_i$

Concept Introduction.

In a projectile motion, the acceleration due to gravity always acts on the vertical component of velocity there is no acceleration on the horizontal component.

 Step 3: Components of the displacement.

The total displacement of the projectile isat an angle.

So,$x$-component of the displacement is:

$d_x=d\cos \varphi$

They-component of the displacement is:

$d_y=d\sin \varphi$

Time is taken by the projectile

The time taken by the projectile is

$\begin{array}{c}t=\frac{d_x}{v_{xi}}\\ =\frac{d\cos \varphi }{v_i\cos {\theta }_i}\end{array}$

They-component of the displacement is also given by

$d_y=v_{yi}t-\frac{1}{2}g{t}^2$…………….(1)

To prove the value of distance.

Solving for the distance d of the projectile up the incline using Equation (1):

$\begin{array}{c}{d}_y=v_{yi}t-\frac{1}{2}g{t}^2\\ d\sin f=v_i\sin {\theta }_i\left(\frac{d\cos f}{v_i\cos {\theta }_i}\right)-\frac{1}{2}g{\left(\frac{d\cos f}{v_i\cos {\theta }_i}\right)}^2\\ 2v_i^{2}d\sin f{\cos }^2{\theta }_i=2v_i^{2}d\cos f\sin {\theta }_i\cos {\theta }_i-g{d}^2{\cos }^2\\ g{d}^2{\cos }^{2}f=2v_i^{2}d\cos {\theta }_i\left\{\cos f\sin {\theta }_i-\sin f\cos {\theta }_i\right\}=2v_i^{2}d\cos \theta \sin \left({\theta }_i-f\right)\\ d=\frac{2v_i^2\cos {\theta }_i\sin \left({\theta }_i-f\right)}{g{\cos }^{2}f}\end{array}$

Hence proved.

Find the initial angle.

(b) To find the initial angle ${\theta }_i$, for which the $d$is maximum.

First, differentiate $d$with${\theta }_i$

$\frac{\mathrm{d}}{{\mathrm{d\theta }}_{\mathrm{i}}}\left(d\right)=\frac{2v_i^{2}d}{g{\cos }^2\varphi }\left[\cos \left({\theta }_i-\varphi \right)\cos {\theta }_i-\sin \left({\theta }_i-\varphi \right)\sin {\theta }_i\right]$

For$d$to maximum,

$\begin{array}{c}\frac{\mathrm{d}}{{\mathrm{d\theta }}_{\mathrm{i}}}\left(d\right)=0\\ \frac{{2\mathrm{v}}_{\mathrm{i}}^2}{{\mathrm{gcos}}^2\mathrm{f}}\left[\cos \left({\mathrm{\theta }}_{\mathrm{i}}-\varphi \right){\mathrm{cos\theta }}_{\mathrm{i}}-\sin \left({\mathrm{\theta }}_{\mathrm{i}}-\varphi \right){\mathrm{sin\theta }}_{\mathrm{i}}\right]=0\\ \frac{{2\mathrm{v}}_{\mathrm{i}}^2}{{\mathrm{gcos}}^2\mathrm{f}}\cos \left(2{\mathrm{\theta }}_{\mathrm{i}}-\varphi \right)=0\mathrm{n}\&\mathrm{ncos}\left(2{\mathrm{\theta }}_{\mathrm{i}}-\varphi \right)=\\ 2{\mathrm{\theta }}_{\mathrm{i}}-\varphi =\frac{\mathrm{\pi }}{2}\\ {\mathrm{\theta }}_{\mathrm{i}}=\frac{\mathrm{\pi }}{4}+\frac{\varphi }{2}\end{array}$

Note of this trigonometric identity:

$\sin a\cos b=\frac{\sin (a+b)+\sin (a-b)}{2}$

Substituting the value of ${\theta }_i=\frac{\pi }{4}+\frac{\varphi }{2}$ to the formula in part (a),

$\begin{array}{c}{\mathrm{d}}_{\max }=\frac{{2\mathrm{v}}_{\mathrm{i}}^2\cos \left(\frac{\mathrm{\pi }}{4}+\frac{\varphi }{2}\right)\sin \left(\left\{\frac{\mathrm{\pi }}{4}+\varphi \right\}-\varphi \right)}{{\mathrm{gcos}}^2\Phi }\\ =\frac{{2\mathrm{v}}_{\mathrm{i}}^2\cos \left(\frac{\mathrm{\pi }}{4}+\frac{\varphi }{2}\right)\sin \left(\frac{\mathrm{\pi }}{4}-\frac{\varphi }{2}\right)}{{\mathrm{gcos}}^2\mathrm{f}}\\ =\frac{{\mathrm{v}}_{\mathrm{i}}^2\left\{\sin \left(\frac{\mathrm{\pi }}{4}-\frac{\varphi }{2}+\frac{\mathrm{\pi }}{4}+\frac{\varphi }{2}\right)+\sin \left(\frac{\mathrm{\pi }}{4}-\frac{\varphi }{2}-\frac{\mathrm{\pi }}{4}-\frac{\varphi }{2}\right)\right\}}{{\mathrm{gcos}}^2\varphi }\\ =\frac{{\mathrm{v}}_{\mathrm{i}}^2(1-\sin \varphi )}{{\mathrm{gcos}}^2\varphi }\end{array}$