- #1
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.) [itex]v_{f}=v_{i}+at[/itex]
2.) [itex]x=x_{o}+v_{ox}\cdot t+\dfrac{1}{2}\cdot a_{x}\cdot t[/itex]
3.) [itex]y=y_{o}+v_{oy}\cdot t+\dfrac{1}{2}\cdot a_{y}\cdot t[/itex]
The Attempt at a Solution
Given:
[itex]v_{o}=18m/s[/itex]
[itex]x_{o}=0m[/itex]
[itex]y_{o}=0m[/itex]
[itex]x_{wall}=20m[/itex]
[itex]v_{ox}=18cos(60°)[/itex]
[itex]v_{oy}=18sin(60°)[/itex]
[itex]a_{x}=0[/itex]
[itex]a_{y}=-9.8m/s^2[/itex]
I solved for the vector components
[itex]v_{ox}=18cos(60°)=15.6m/s[/itex]
[itex]v_{oy}=18sin(60°)=9m/s[/itex]
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.
Attachments
Last edited by a moderator: