Motion in Space: Calculus 3 prob::confused.

[Joker2k999]

Good Day,
I can't understand why i am not getting the right answer for this _Motion in Space problem.

A gun has a muzzle speed of 80 m/s. What angle of elevation should be used to hit an object 180 meters away?

theta = sin^-1 (180g/d^2) right?

so....

theta = sin^-1((180 * 9.8)/(180^2))

theta = 16.26 deg.
= 0.283 rad.
Right?
can't understand how this isn't right??

Help is appreciated.
Thx in adv.
JkR.

 

[tkhunny]

Something about \(\displaystyle \frac{180}{180^{2}}\) doesn't look right.

You may also wish to correct your formula from "\(\displaystyle \theta\)" to "\(\displaystyle 2\theta\)"

 

[soroban]

Hello, Joker2k999!

A gun has a muzzle speed of 80 m/s.
What angle of elevation should be used to hit an object 180 meters away?

We are expected to be familiar with the Trajectory Functions.

Horizontal displacement: .\(\displaystyle x \;=\;(v_o\cos\theta)t\)
Vertical displacement: .\(\displaystyle y \;=\;h_o + (v_o\sin\theta)t - 4.9t^2\)

. . where: .\(\displaystyle \begin{Bmatrix}v_o &=& \text{initial velocity} \\ \theta &=& \text{elevation angle} \\ h_o &=& \text{initial height} \end{Bmatrix}\)


I will assume that the gun and the target are at ground level.

. . Then we have: .\(\displaystyle \begin{Bmatrix}x \:=\: (80\cos\theta)t \\ y \:=\: (80\sin\theta)t - 4.9t^2 \end{Bmatrix}\)


We want the condition for which: .\(\displaystyle x = 180,\;y = 0\)

. . So we have: .\(\displaystyle \begin{Bmatrix} (80\cos\theta)t \:=\:180 & [1] \\ (80\sin\theta)t - 4.9t^2 \:=\:0 & [2] \end{Bmatrix}\)


From [2]: .\(\displaystyle t(80\sin\theta - 4.9t) \;=\;0\)

And we have: .\(\displaystyle \begin{Bmatrix} t \:=\:0 \\ t \:=\:\frac{80\sin\theta}{4.9} \end{Bmatrix}\)

Substitute into [1]: .\(\displaystyle (80\cos\theta)\left(\dfrac{80\sin\theta}{4.9}\right) \:=\:180 \quad\Rightarrow\quad 6400\sin\theta\cos\theta \:=\:4.9(180)\)

. . . \(\displaystyle 3200(2\sin\theta\cos\theta) \:=\:882 \qquad\Rightarrow\qquad 3200\sin2\theta \:=\:882\)

. . . \(\displaystyle \sin2\theta \:=\:\dfrac{882}{3200} \:=\:0.285625 \quad\Rightarrow\quad 2\theta \:=\:15.99926354^o \:\approx\:16^o\)


Therefore: .\(\displaystyle \theta \:\approx\:8^o\)

 

[Joker2k999]

Cool, thx guyz.

yea, sorry i didn't include the 2Theta. i knew i was gonna have to divide it out anyway. Thx for the explanation soroban! Yea i tried that. But my Professor said i had made an error... i'll take it to him again Monday.

Thx guys.

Cheers.
JkR.

 

[tkhunny]

i didn't include the 2Theta. i knew i was gonna have to divide it out anyway.

That's make absolutely no sense. Please make sure you know what you are doing. Don't make up new algebra until you get a better grip on old algebra.