How Long Does It Take for a Launched Ball to Hit the Wall?

[Huski]

Homework Statement

*see attachment*[/B]
A ball is launched from (0,0) with an initial speed of 18.0 m/s at a 60 degree angle (as shown above) above the horizontal. The ball starts at the floor and 20.0 m left of a wall. Neglect air resistance. The ball hits the wall at some time.

A. Calculate when the ball hits the wall.

Homework Equations

[/B]
1.) $v_{f}=v_{i}+at$

2.) $x=x_{o}+v_{ox}\cdot t+\dfrac{1}{2}\cdot a_{x}\cdot t$

3.) $y=y_{o}+v_{oy}\cdot t+\dfrac{1}{2}\cdot a_{y}\cdot t$

The Attempt at a Solution

Given:
$v_{o}=18m/s$
$x_{o}=0m$
$y_{o}=0m$
$x_{wall}=20m$
$v_{ox}=18cos(60°)$
$v_{oy}=18sin(60°)$
$a_{x}=0$
$a_{y}=-9.8m/s^2$

I solved for the vector components
$v_{ox}=18cos(60°)=15.6m/s$

$v_{oy}=18sin(60°)=9m/s$

I don't know my y component of how high the ball will go and if I solve for the y-component (equation 3), I have all the variables except for the 'y' on the left hand of the equation. I need to solve for 't' (time), but I don't know how to get y first? The y on the left side is the final height. Usually, I deal with problems when the ball hits the ground and I can set y = 0 (since 0 is the ground). Here, the ball hits the wall and we are not sure what y (the final height) is equal to, any hints? Thank you.

 

[FactChecker]

None of the numbers in your equations match the statement problem or the diagram (the statement and diagram don't match each other). Please present one consistent problem and show your calculations for that problem.

 

[Huski]

That was an accident. I had two different problems and I copied the wrong one, let me edit that.

 

[haruspex]

18cos(60°)=15.6m/s

Try that again. And the v_y.

Your equations 2 and 3 seem to be missing something at the end.

Try working with the x direction first.

 

[Huski]

Oh, I need to solve for x, okay thanks.

$x=x_{o}+v_{ox}t+\dfrac{1}{2}(0)t^2$

$20=0+9t+\dfrac{1}{2}(0)t^2$

$20=9t$

$t=2.22 seconds$

I think this will give me the time the ball hits the wall? So we don't care about the y-direction?
Also, I forgot to square the two t in the 2 and 3 equations. Thanks for that.

Fixes:
$18\cos(60)=9m/s$

$18\sin(60)=15.6m/s$

$x=x_{o}+v_{ox}\cdot t+\dfrac{1}{2}\cdot a_{x}\cdot t^2$

$y=y_{o}+v_{oy}\cdot t+\dfrac{1}{2}\cdot a_{y}\cdot t^2$