# Projectile - what angle?

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?

 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #2 Feb 19, 2007, 11:34 PM
What is the "nearest edge"?
 jeepey Posts: 19, Reputation: 1 New Member #3 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 Posts: 5,319, Reputation: 3601 Uber Member #4 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 Posts: 19, Reputation: 1 New Member #5 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 Posts: 5,319, Reputation: 3601 Uber Member #6 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 Posts: 19, Reputation: 1 New Member #7 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 Posts: 5,319, Reputation: 3601 Uber Member #8 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 Posts: 19, Reputation: 1 New Member #9 Feb 27, 2007, 07:10 AM
Not sure if I'm close, I got 67 deg for the angle?
 Capuchin Posts: 5,319, Reputation: 3601 Uber Member #10 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.

## Check out some similar questions!

Projectile motion problem [ 5 Answers ]

A place kicker must kick a football from a point 36.0 m from the goal, and the ball must clear the crossbar, which is 3.05 m high. When kicked, the ball leaves the ground with a speed of 20.0m/s at an angle of 53 to the horizontal. A) By how much does the ball clear or fall short of clearing the...

Zero Launch Angle [ 1 Answers ]

A diver runs horizontally off the end of a diving board with an initial speed of 1.75m/s. If the diving board is 3m above the water, what is the diver's speed just before he enters the water? When I worked out this problem, I came up with an answer of 7.67m/s. The answer at the back of the...

Angle stop [ 3 Answers ]

What is a standard 1/2" compression angle stop? I am installing a sink faucet supply line