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

What is the "nearest edge"?

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

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

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

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 :)

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.

-9.025 is from

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

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

right?

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

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

I'm bad at this part of math. your not the only one! Lol

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

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

OK. I follow to
0.7%20=%20\frac{19sin\theta}{cos\theta}%20-%209.025cos^2\theta
the real problem is solving for \theta... obviously
sin/cos=tan and
\cos^2 \theta = \frac {1 + \cos (2\theta ) } 2 \\
\text {so} \\
0.7=19 \tan \theta - 4.5125 - 4.5125 \cos (2\theta) \\
5.2125= 19 \tan \theta - 4.5125 \cos (2\theta)


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 :)

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.

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?

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