A ball thrown over a house(Projectile Motion)

[cde42003]

Can anyone help me out with the second half of this problem? I did it awhile ago and as I am trying to review for a test I can't remember how I solved it.

You're 6.0 m from one wall of a house. You want to toss a ball to your friend who is 6.0 m from the opposite wall. The throw and catch each occur 1.0 m above the ground.

A.What minimum speed will allow the ball to clear the roof? I know this is 13.3 m/s

B. At what angle should you toss the ball? I can't seem to figure this part out now. I am sure it is not that difficult but I am stumped. Can anyone refresh my memory? Thanks

 

[robphy]

Short answer: use the range formula. [But only use it if you understand how to derive it.]

 

[HallsofIvy]

First you will need to calculate the height of the house- then subtract 1 to allow fo the initial and final heights (are you kneeling while throwing the ball?) and the horizontal distance between yourself and your friend. Now, use the height and distance (x and y) components so the the maximum height of the ball is at least as large as the height of the house and the horizontal distance is equal to the distance between you and your friend.

 

[robphy]

On topic, but probably not what the original poster wants...
http://www.du.edu/~jcalvert/math/parabola.htm .
The first figure suggests an interesting geometric interpretation of the answer to part B [at what angle should the ball be thrown to just clear the roof].

 

[cde42003]

Thanks, I figured it out now.