roof toss

bricker9236

1. The problem statement, all variables and given/known data
A ball is thrown OFF a roof at a height 35.0m above ground. The ball hits the ground 29m from the base of the building. the max height of the balls trajectory is 12 m above the height from which the thrower released it.

a) find the initial vertical component of the velocity.
b) find the total time that the ball is in flight (from the time it is thrown until it lands on the ground)
c) find the horizontal compoent of the initial velocity required for the ball to land 29 m away from the building.
d) at what angle is the ball thrown from the roof? (measure angle above horizontal =0 degrees)


2. Relevant equations
Not sure, I was just randomly trying to figure it out and this problem frustrated me.



3. The attempt at a solution
a) 10.9 .. that is the answer i got, if anyone could confirm or help me out.
b) 2.22 seconds
c) 14.23 m/s not sure, again
d) 40 degrees again, not very sure .. i used tan inverse of 12/29 wasnt sure if it shiould be tan inverse of (35+12)/29 not sure

rock.freak667

So let's use these

[tex]y=y_0 +ut -\frac{1}{2}gt^2[/tex]

[tex]v^2=u^2-2g(y-y_0)[/tex]

[tex]v=u+at[/tex]

u=initial velocity
v=final velocity
y₀=initial height.

Can't really check your answers if you yourself don't know how you got them.


Now we start, let's call the vertical component of velocity v_y and the horizontal v_x

We are given the maximum height as 12m from the height thrown. What do you know about the final vertical velocity at the maximum height?

bricker9236

the final vertical velocity at max height would be 0 bc at any max height the velocity is always 0

rock.freak667

Right, so using v_y=0, in our possible equations containing v, we can get

[tex]0=u^2-2g(y-y_0)[/tex]


or

[tex]0=u+at[/tex]


Which would you like to use or which do you think is better to use?