# Kinematics Issue w/ negative Velocity [closed]

Given: $$Y = v\sin(\theta)t-\frac 12 gt^2$$ $$D=v\cos(\theta)t$$

Thus: $$\frac{D}{v\cos(\theta)} = t$$

Substituting we get: $$Y=v\sin(\theta)\frac{D}{v\cos(\theta)}-5\left( \frac{D}{v\cos(\theta)}\right)^2$$

Reducing we get: $$Y=\sin(\theta)\frac{D}{\cos(\theta)}-5\frac{D^2}{v^2\cos(\theta)^2}$$ Reducing we get: $$Y+\frac{5(D)^2}{v^2\cos^2(\theta)} =\sin(\theta)\frac{D}{\cos(\theta)}$$ Reducing we get: $$\frac{1}{v^2\cos^2(\theta)} =\sin(\theta)\frac{D}{5D^2\cos(\theta)}-\frac{Y}{5D^2}$$

Reducing we get: $$v^2=\frac{5D}{\sin(\theta)\cos(\theta)}-\frac{5D^2}{Y\cos^2(\theta)}$$

Thus $$v=\sqrt{\frac{5D}{\sin(\theta)\cos(\theta)}-\frac{5D^2}{Y\cos^2(\theta)}}$$

Sorry about the formatting, I'm not used to this. Feel free to ask any questions about the problem. Basically I'm trying to get a velocity without ever having to know t. FYI the above equation doesn't work because it somehow sometimes produces negative velocity. PLS help.

Edit: Commenting about the positive/negative nature of square roots isn't helpful because the issue is with values like D = 7, $$\theta =$$ 45.

## closed as off-topic by Aaron Stevens, user191954, Qmechanic♦Sep 28 '18 at 5:40

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Community, Qmechanic
If this question can be reworded to fit the rules in the help center, please edit the question.

• I tried to edit the equations as best as I could. You need to format to entire equations with MathJax, not just certain portions. Please check the equations to make sure I did what you originally had. It was hard to follow all of the nested parentheses, especially on my phone. Then others can easily look through your work. I see some errors I think, but I'm not sure if that is my fault. If there is an error, hopefully you can see what I did with the formatting to fix the errors yourself. – Aaron Stevens Sep 28 '18 at 2:28
• Hi welcome to stackexchange. We don't usually accept "check-my-work" style questions. We focus on questions about physical concepts. Is there a particular concept you are having trouble with? – Paul T. Sep 28 '18 at 2:33
• Yeah there was one little thing with the distribution of squares, I'll touch it up. Thank you so much! @Aaron – Christheyankee Sep 28 '18 at 2:56
• @Paul I'm not sure it's a concept, but how are we getting negative velocities for angles such as 45 when D is 7? – Christheyankee Sep 28 '18 at 3:00
• The line before the square root is wrong. $\left(\frac ab + \frac cd\right)^{-1}\neq \frac ba +\frac dc$ – Aaron Stevens Sep 28 '18 at 3:35

You are correct until this line, after which you did not invert the fraction properly: $$\frac{1}{v^2\cos^2\theta} =\sin\theta\frac{D}{5D^2\cos\theta}-\frac{Y}{5D^2} = \frac{D \sin\theta - Y \cos\theta}{5D^2 \cos\theta}$$ So $$v^2\cos^2\theta = \frac{5D^2 \cos\theta}{D \sin\theta - Y \cos\theta}$$ $$v^2 = \frac{5D^2}{D \sin\theta \cos\theta - Y \cos^2\theta}$$
• @Christheyankee: That shouldn't happen. The values of $Y$, $D$ and $\theta$ are not really independent. For instance, we can't have a near-vertical launch and yet have $D$ > $Y$. Try plugging in some real data. – user7777777 Sep 28 '18 at 11:42