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M*-*Fares

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posted by M*-*Fares on 11/5/2009 10:54:36 PM | status: Closed | Earned Karma: 51

Calc lll

Course	Textbook	Chapter	Problem	Needs by
Calculus	N/A	N/A	N/A	11/6/2009 at 4:00:00 PM

Question Details:

I need help in detail, please

If a shell is fired from ground level with a speed of 160 ft/sec and elevation is 30^o, determine:

a) The vector valued function for the object’s path

b) The Maximum height of the object

c) Horizontal distance of the object

d) The speed of the object at impact

Tags: Calculus, Calculus, Calculus

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Answers:

anonymousx...

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posted by anonymousxyz on 11/6/2009 1:17:47 AM | status: Live

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Response Details:

$\text{(a)}$

$\text{(i)}$

$\text{Consider the horizontal component of the shell's motion:}$

$\begin{align*} & a_x(t) = 0 \text{ ft/s}^2 \\\\& v_x(t) = \int a_x(t) \;\text{d}t = C \end{align*} \\\\$

$\begin{align*} & v_x(0) = 160 \cos 30^\circ \text{ ft/s }&\Rightarrow\\\\& C = 160 \cos 30^\circ &\Rightarrow\\\\& v_x(t) = 160 \cos 30^\circ \end{align*}$

$\begin{align*} & x(t) = \int v_x(t)\;\text{d}t = ( 160 \cos 30^\circ )t + C \end{align*}$

$\begin{align*} & x(0) = 0 \text{ ft} &\Rightarrow\\\\& C = 0 &\Rightarrow\\\\& x( t ) = ( 160 \cos 30^\circ )t \end{align*}$

$\text{(ii)}$

$\text{Consider the vertical component of the shell's motion:}$

$\begin{align*} & a_y(t) = -32 \text{ ft/s}^2 \\\\& v_y(t) = \int a_y(t) \;\text{d}t = -32t + C \end{align*} \\\\$

$\begin{align*} & v_y(0) = 160 \sin 30^\circ \text{ ft/s }&\Rightarrow\\\\& C = 160 \sin 30^\circ &\Rightarrow\\\\& v_y(t) = -32t + 160 \sin 30^\circ \end{align*}$

$\begin{align*} & y(t) = \int v_y(t)\;\text{d}t = -16t^2 + ( 160 \sin 30^\circ )t + C \end{align*}$

$\begin{align*} & y(0) = 0 \text{ ft} &\Rightarrow\\\\& C = 0 &\Rightarrow\\\\& y( t ) = -16t^2 + ( 160 \sin 30^\circ )t \end{align*}$

$\text{(iii)}$

$\mathbf{r}(t) = \big<\;x(t), y( t) \;\big> = \big<\;(160 \cos 30^\circ)t, -16t^2 + (160 \sin 30^\circ)t \;\big>$

$\text{(b)}$

$\begin{align*} & v_y(t) = 0 &\Rightarrow\\\\& {-32t} + 160 \sin 30^\circ = 0 &\Rightarrow\\\\& {32t} = 160 \sin 30^\circ &\Rightarrow\\\\& t = 5 \sin 30^\circ \Rightarrow\\\\& t = 2.5 \text{ s} \end{align*}$

$y\left( 2.5\right) = -16( 2.5 )^2 + (160 \sin 30^\circ)( 2.5 ) = -100 + 200 = 100 \text{ ft }$

$\text{(c)}$

$\begin{align*} & y(t) = 0 &\Rightarrow\\\\& {-16}t^2 + ( 160 \sin 30^\circ )t = 0 &\Rightarrow\\\\& t( -16t + 160 \sin 30^\circ ) = 0 &\Rightarrow\\\\& t( -16t + 80 ) = 0 &\Rightarrow\\\\& t = 0 \text{ s} \;\;\text{or}\;\; t = 5 \text{ s} \end{align*}$

$x\left( 5 \right) = (160 \cos 30^\circ)(5) = 400\sqrt{3} \text{ ft }$

$\text{ (d) }$

$\mathbf{r}'(t) = \big<\;160 \cos 30^\circ, -32t + 160 \sin 30^\circ \;\big>$

$| \mathbf{r}'(t) | &= \sqrt{ ( 160 \cos 30^\circ )^2 + ( -32t + 160 \sin 30^\circ )^2 }$

$| \mathbf{r}'(5) | &= \sqrt{ ( 160 \cos 30^\circ )^2 + ( -32(5) + 160 \sin 30^\circ )^2 } = \sqrt{25600} = 160 \text{ ft/s }$

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