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Quick Questions: Antiderivative of 1/sec^2(x) and Acceleration Negative When Throwing a Ball Up?

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 (x^2/sqrt(x)) + (1/sin^2(x))?

I can got:
((2/5)x^(5/2)) + (1/tan(x)) + c
But I am pretty sure (1/tan(x)) isn't right.
2 Solutions
Since the acceleration is constant, it will be the same while the ball is going up as it is while the ball is coming down.

1/tan(x) is right
You can decide wheather the acceleration is to be + or -, but you have to be consistant throughout the problem. Most people will say that after a ball is thrown the accelleration is negative. That is the acceleration due to gravity, with most people calling "down" -. Now when you throw a ball up, you are giving it a positive acceleration while it is in your hand going from rest to some velocity (49 m/s possibly). But as soon as it leaves your had the acceteration will be 32 feet /sec/sec downwards due to gravity (on the earth anyway).
The accelaration is that experinced by the ball after it has been thrown and is that of gravity ie negative. Regardless of what the direction the ball is thrown the accelaration is negative ie downwards.

integral( 1/sin^2x )dx =integral( csc^2x )dx = - cot x + C
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The ball is slowing, so negative acceleration
You've got the definition of secant wrong.  secant is 1/cos, not 1/sin.  See http://mathworld.wolfram.com/Secant.html
cosecant is 1/sin.

The integral of 1/sin^2(x) is -cot(x) + c  or -1/tan(x) + c
The integral of 1/cos^2(x) is tan(x) + c.

Yes, I droped a sign, sorry.
Acceleration is the rate of change of velocity with respect to time. Velocity is a vector and so has a directional component.
The acceleration can be thought of as either positive or negative depending on whether the initial velocity is thought of as positive or negative.
If the initial velocity is negative then the acceleration is positive. If the initial velocity is positive then the acceleration is negative.
The choice of whether the initial velocity of the ball is positive or negative is entirely up to the author though it is standard to say that acceleration due to gravity is (positive) 9.8 m/s² so if the ball was thrown upwards the initial velocity is negative.

Just in case you're interested gravity at the North Pole is 9.832 m/s², Java (near the equator) it's 9.782 m/s².


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