# Kinematics Projectile Motion Problem

1. Feb 6, 2015

### joeyoung1996

1. The problem statement, all variables and given/known data
A ball launched from ground level lands 2.85s later on a level field 48 m away from the launch point. Find the magnitude of the initial velocity vector and the angle it is above the horizontal. (Ignore any effects due to air resistance.)

2. Relevant equations
$x=x_0+v_{0x}t+\frac{1}{2}at^2$
$y=y_0+v_{0y}t+\frac{1}{2}at^2$

3. The attempt at a solution
So, I worked out this problem seemingly without an issue, but WebAssign keeps telling me the answers are wrong, so maybe a bit of new eyes can point out what I've done wrong.

I start by attempting to solve for the x component of the velocity vector:
$x=x_0+v_{0x}t+\frac{1}{2}at^2$
$48m=0m+v_{0x}(2.85s)+\frac{1}{2}(0)(2.85s)^2$
$48m=v_{0x}(2.85s)$
$v_{0x}=\frac{48m}{2.85s}$
$v_{0x}=16.84\frac{m}{s}$
I then found the y component of the velocity:

$y=y_0+v_{0y}t+\frac{1}{2}at^2$
$0m=0m+v_{0y}(2.85s)+\frac{1}{2}(-9.8\frac{m}{s^2})(2.85s)^2$
$0=v_{0y}(2.85s)-13.96m$
$v_{0y}=\frac{13.96m}{2.85s}$
$v_{0y}=4.9\frac{m}{s}$
Now that I have both components, I use the Pythagorean Theorem to find the magnitude of the velocity:

$||\vec{v_0}|| = \sqrt{v_{0x}^2+v_{0y}^2}$
$||\vec{v_0}||= \sqrt{(16.84\frac{m}{s})^2+(4.9\frac{m}{s})^2}$
$||\vec{v_0}||=17.5\frac{m}{s}$
Finally, I use these to find the angle at which the ball was fired:

$sinx=\frac{4.9}{17.5}$
$x=sin^{-1}(\frac{4.9\frac{m}{s}}{17.5\frac{m}{s}})$
$x=16.22^{\circ}$
I figured I have just made a simple, stupid mistake, but for the life of me, I cannot find it.

Last edited: Feb 6, 2015
2. Feb 6, 2015

### TSny

Did you forget to square the time in the last term above? Otherwise, your work looks good to me.

3. Feb 6, 2015

### joeyoung1996

LOL I didn't square the time haha!! I feel silly now; thank you so much!!

4. Feb 6, 2015

### TSny

You're welcome. It was just a careless error, like we all make. Hope it works out now.