Two-Dimensional Projectile Motion: Finding Vertical and Horizontal Distances

[philok]

Homework Statement

a ball is thrown up to the top of a building 6 m tall. It is thrown at a 53 degree angle above the horizontal at a point 24m from the base of the building. The ball takes 2.20s to reach a point vertically above the wall.
a) find the vertical distance by which the ball clears the wall.
b) find the distance from the wall to the point on the roof where the ball lands

Homework Equations

Kinematics and other trig equations.

The Attempt at a Solution

I know that V_x is 18.1 m/s and V_y is 3.42 m/s. From there I am not sure where to go. I have tried several different things and I am not coming up with the answer. Please tell me how to go about this problem to see if I am headed in the right direction.

Thanks.

 

[thebigstar25]

are you given the initial velocity when the ball thrown?

 

[philok]

Sorry about that. No It does give it. So that was my first step.
in the x- direction the intial velocity is 18.1 m/s and the y direction is 3.42.

 

[philok]

That was suppose to be *doesn't give it.

 

[thebigstar25]

assuming that you got Vx and Vy correctly, part (a) is asking for the maximum vertical height ,hint : what happens when the ball reaches maximum height? ( you can i.e use this equation y = voyt + 0.5gt^2 , g here as a vector quantity) ..

and part(b) is asking for the distance from the wall to the point on the roof where the ball lands, you can start this part by first finding the range of the travel (max. horizontal distance) and whatever you get from there subtract it the 24m from it .. can you go from there?