## Positive or Negative Acceleration? and Antiderivative of 1/(sin^2)?

I have two quick questions. The first is, if a ball is thrown from a bridge 25 meter above the ground at 49m/s, is the acceleration negative? I thought since it was thrown up, it would be positive, but the book has a similar problem, a ball is thrown upward at a speed of 48ft/s from a cliff 432ft above the ground, and it says the acceleration is -32.

Second, what is the antiderivative of $$\frac{x^2}{\sqrt{x}} + \frac{1}{sin^2x}$$
I can got:
$$\frac{2}{5}x^{5/2}+\frac{1}{\tan{x}}+c$$

But I am pretty sure $$\frac{1}{\tan{x}}$$ isn't right.
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 Recognitions: Gold Member Homework Help Science Advisor As for the acceleration question: Ask yourself: 1. WHAT FORCES ACT UPON THE BALL ONCE IT HAS LEFT YOUR HAND? What is that force's direction? (Assuming there is only one force acting on the ball) 2. How is a force related to acceleration? As for the derivative: You're right, $$\frac{d}{dx}\frac{1}{tan(x)}=-\frac{1}{\sin^{2}x}$$ , not $$\frac{1}{\sin^{2}x}$$
 Gravity will always pull it down, no matter if it was thrown up or down, so acceleration is always negative (assumming it's on Earth, under normal conditions)? The antiderivative is: $$\frac{2}{5}x^{5/2}-\frac{1}{\tan{x}}+c$$ The only thing wrong was the sign?

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## Positive or Negative Acceleration? and Antiderivative of 1/(sin^2)?

Yes, right on both problems.
Note that you will most often find the multiplicative inverse of the tangens function called cotangens:
$$cot(x)\equiv\frac{1}{tan(x)}$$
 I just thought of something else. $$If \csc{x} = \frac{1}{\sin{x}} , then \csc^{2}x = \frac{1}{\sin^{2}x}$$ The antiderivative of $$\csc^{2}{x}$$ is $$-\cot{x}+c$$ Making the final answer be: $$\frac{2}{5}x^{5/2}-\cot{x}+c$$ and not $$\frac{2}{5}x^{5/2}-\frac{1}{\tan{x}}+c$$
 Recognitions: Gold Member Homework Help Science Advisor As I wrote, cotangens IS the multiplicative inverse of tangens. Both answers are equally valid.

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