Antiderivatives/derivatives

**MathNeedy18** · November 18th 2007, 10:14 PM

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

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.

**topsquark** · November 19th 2007, 06:17 AM

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

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

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