Projectile - what angle?

Capuchin · Feb 19, 2007, 11:34 PM

What is the "nearest edge"?

jeepey · Feb 19, 2007, 11:38 PM

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

Capuchin · Feb 19, 2007, 11:50 PM

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

$V_v = 14sin heta$

$V_h = 14cos heta$

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 you, a, s and t from above:

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

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

Now substituting in $V_v$ and $V_h$

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

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

Solve for $heta$

jeepey · Feb 20, 2007, 12:08 AM

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

Capuchin · Feb 20, 2007, 12:12 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 26, 2007, 11:17 PM

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 26, 2007, 11:28 PM

-9.025 is from

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

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

Right?

jeepey · Feb 27, 2007, 07:10 AM

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

Capuchin · Feb 27, 2007, 07:38 AM

I'm not too sure either

I'm bad at this part of math. I'll have a look at it later tonight.

Projectile - what angle?

Add your answer here.

Check out some similar questions!