# Antiderivatives/derivatives

• Nov 18th 2007, 11:14 PM
MathNeedy18
Antiderivatives/derivatives
I do not know what to do with the 45 degrees!

A ball is shot at an angle of 45 degrees into the air with initial velocity of http://webwork1.math.utah.edu/math12...109294img1.gif ft/sec. Assuming no air resistance, how high does it go?

How far away does it land?
Hint: The acceleration due to gravity is 32 ft per second squared.
• Nov 19th 2007, 07:17 AM
topsquark
Quote:

Originally Posted by MathNeedy18
I do not know what to do with the 45 degrees!

A ball is shot at an angle of 45 degrees into the air with initial velocity of http://webwork1.math.utah.edu/math12...109294img1.gif ft/sec. Assuming no air resistance, how high does it go?

How far away does it land?
Hint: The acceleration due to gravity is 32 ft per second squared.

I only have a moment, so I'll be brief.

The motion in perpendicular directions is independent of each other. So we have:
$x = x_0 + v_{0x}t + \frac{1}{2}a_xt^2$

$y = y_0 + v_{0y}t + \frac{1}{2}a_yt^2$

Here $a_x = 0$ because nothing is making it accelerate in the x direction and $a_y = -g$.

Now, you have an initial velocity $v_0$ pointed at 45 degrees above the horizontal. This is a vector. So your components in the x and y direction will depend on this angle.

For example, let's call +y upward and +x along the ground in the relative direction the ball was shot in. Then
$v_{0x} = v_0~cos(45^o)$
and
$v_{0y} = v_0~sin(45^o)$

-Dan