# Throwing a Ball and determining possible angles.

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1. Jan 16, 2016

### CGI

• HW Template missing as it was moved from another forum
A man throws a ball horizontally with an initial speed of 20 m/s. Determine the two possible angles theta 1 and theta 2 of release so that the ball strikes point B. The distance between the man and point B is 15 meters. The initial height of the ball is 1.25 meters.

Theta 1 = 5.8 Degrees
Theta 2 = 79.4 Degrees

I know the answer, but I'm not sure how to get there. I thought about the range equation, but it doesn't seem to be working out. Any help would be much appreciated!

What I have so far:

15 = 20cos(Θ)t so, t = 15/(20cosΘ)

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2. Jan 16, 2016

### Student100

Is the 20 m/s only in the x direction, or is it the total velocity? The way you wrote it makes me think it's the first, the picture makes me think it might be the second.

3. Jan 16, 2016

### CGI

I'm sorry, I meant that the 20 m/s was the total velocity

4. Jan 16, 2016

### Student100

Okay, whats wrong with the range equation? Why do you think it isn't going to work?

5. Jan 16, 2016

### CGI

So when I try the range equation, I don't get the right answer.
I tried:

15 = (20)^2sin(2theta)/g

And I get that theta = 10.79°

6. Jan 16, 2016

### Student100

Why doesn't it work though? What piece of information is being left out?

7. Jan 16, 2016

### CGI

I'm thinking the initial height is what's left, but I'm not sure how I would use that. I did think about how when the ball reaches point B, the height will equal 0.

8. Jan 16, 2016

### Student100

Yes, the equation you used is telling you that launching the ball at 10 some degrees will ensure the ball is back at 1.25 meters when the x distance is 15 meters. So the initial height here is important. Do you know another way to find the distance of landing that takes into account the initial height of launch?

9. Jan 16, 2016

### CGI

Would I use the height of a projectile equation?

h = initial height + (Vo)^2sin^2(theta)/2g

And set height equal to zero and solve for theta? Or would that still be incorrect?

10. Jan 16, 2016

### Student100

Have you ever seen the derivation for: $$\theta = tan^{-1}(\frac{v^2 \pm \sqrt{v^4 - g(gx^2+2yv^2)}}{gx})$$

11. Jan 16, 2016

### CGI

Oh, I've never actually seen that before. What is that equation called? And how does one use it?

12. Jan 16, 2016

### Student100

x and y are your target coordinates, in your case, x = 15 and y = -1.25. The $\pm$ will give you both angles, and your answer. Answer isn't important, lets look at what we're actually doing.

First at how to derive it: start with your two kinematic equations for x and y in relation to simple projectile motion when x and y equal 0: $$x = vtcos(\theta)$$ $$y = vtsin(\theta) -\frac{1}{2}gt^2$$

Solve x for time and sub into y.

13. Jan 16, 2016

### CGI

Wow! Okay that makes a lot of sense and I actually got the right answer. Thank you so much for your help!