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jeepey · Feb 20, 2007, 12:14 AM

Hello All,

I have a question on projectiles, but I am unsure what formula I need to use to calculate the required angle.

A projectile leaves the nozzle at 14 m/s, at what angle should the nozzle be so that it hits the nearest edge at the top of a wall 0.7m high with a horizontal distance oy 19m from the nozzle?

Thanks in advance

Capuchin · Feb 20, 2007, 12:34 AM

What is the "nearest edge"?

jeepey · Feb 20, 2007, 12:38 AM

as in the top of the wall. - so the point x= 19m and y=0.7m from the nozzle

Capuchin · Feb 20, 2007, 12:50 AM

Okay, so we need equations for $V_v$ denoting the vertical initial velocity and $V_h$ denoting the initial horizontal velocity.

Using trigonometry we can easily see that

$V_v = 14sin\theta$

$V_h = 14cos\theta$

Now, we need to work out the time that the ball is in the air, this is easier to do with the horizontal motion, as there is no acceleration.
Time before the ball hits the wall in the horzontal direction:

$t = \frac{d}{v}$

$t = \frac{19}{V_h}$

Now, we know how high the ball needs to be at this point, so we can look at the vertical motion.
Using one of the suvat equations:

$s = ut + \frac{1}{2}at^2$

substituting in u, a, s and t from above:

$0.7 = V_v\frac{19}{V_h} + \frac{1}{2}*(-9.8)(\frac{19}{V_h})^2$

$0.7 = \frac{19V_v}{V_h} + \frac{(-9.8)*19^2}{2V_h^2}$

Now substituting in $V_v$ and $V_h$

$0.7 = \frac{19*14sin\theta}{14cos\theta} + \frac{(-9.8)*19^2}{2(14cos\theta)^2}$

$0.7 = \frac{19sin\theta}{cos\theta} - 9.025cos^2\theta$

Solve for $\theta$

jeepey · Feb 20, 2007, 01:08 AM

Many thanks for the quick reply. I'll post up my answer for check later if that's OK.

Capuchin · Feb 20, 2007, 01:12 AM

Sure :)

I think there must be an easier way to do it because this seems overly complicated. We'll see :)

Make sure I make sense! :)

Okay I edited the first post, i forgot the $\frac{19}$ for t

now it reads $0.7 = \frac{19sin\theta}{cos\theta} + 9.025cos^2\theta$

Please check my working, i'm only human, and i'm doing this all in my head.

Another problem, gravity is in the negative direction:

$0.7 = \frac{19sin\theta}{cos\theta} - 9.025cos^2\theta$

try that :)

jeepey · Feb 27, 2007, 12:17 AM

Hello,
I am just revisiting this question on the projectile angle, I am a little confused as to how the -9.025cos2Ø is derived from the previous step.

Sorry if this seems a daft question.

Capuchin · Feb 27, 2007, 12:28 AM

-9.025 is from

$\frac{(-9.8)*19^2}{2(14cos\theta)^2}$

$\frac{(-9.8)*19^2}{2(14)^2} = -9.025$

right?

jeepey · Feb 27, 2007, 08:10 AM

Not sure if I'm close, I got 67 deg for the angle?

Capuchin · Feb 27, 2007, 08:38 AM

I'm not too sure either :D I'm bad at this part of math. I'll have a look at it later tonight.

jeepey · Feb 27, 2007, 08:41 AM

I'm bad at this part of math.

your not the only one! Lol

Capuchin · Feb 27, 2007, 08:46 AM

Hopefully asterisk will pop his head in, I told him to, he'll hopefully have some time to help right now :)

asterisk_man · Feb 27, 2007, 09:21 AM

somewhat busy right now so don't hold me to this!

OK. I follow to
$0.7 = \frac{19sin\theta}{cos\theta} - 9.025cos^2\theta$
the real problem is solving for $\theta$ ... obviously
sin/cos=tan and
$\cos^2 \theta = \frac {1 + \cos (2\theta ) } 2 \\<br /> \text {so} \\<br /> 0.7=19 \tan \theta - 4.5125 - 4.5125 \cos (2\theta) \\<br /> 5.2125= 19 \tan \theta - 4.5125 \cos (2\theta) <br />$

unfortunately my brain is unable to figure out how you solve for theta at the moment
numerically i think the answer is:
0.41113094848254 radians = 23.5560682 degrees

if i substitute into the first equation i listed then i do get 0.7

someone else can try to substitute this back into the original equations to see if it really does provide the correct answer. Or I'll try more later if i get a chance.

sorry i couldn't provide more help. I also dislike trig identities :)

jeepey · Feb 27, 2007, 11:46 PM

Thanks for the help guys, this problem is more complicated than 1st seems. Some parts of the workings I am struggling to follow, but have helped.

jeepey · Mar 5, 2007, 06:24 AM

sorry I am re visiting this question

I'm not sure if I'm right, but would the -9.025cos^2 theta not be -9.025cos^-2 theta as its being divided by?

Capuchin · Mar 5, 2007, 06:30 AM

I don't think it is being divided by..