# Launch angle and Projectiles

## Homework Statement

Suppose a very narrow, very tall wall of height H stands between two children who are trying to play catch. The wall is a distance x[wall] from the thrower, and the catcher is distance X[F] from the thrower. In this problem you will calculate the lunch angle (theta) from the horizontal that the thrower must use to just clear the wall and still land at the catcher. You may neglect the height of the children and the width of the wall.
a)Draw a diagram of the problem. (hint: do not assume that the wall's position is halfway between, or that the ball's peak height occurs at the wall)
b0 What is the vertical position of the ball (as a function of time), expressed in terms of the V and the launch angle (theta)?
c) What is the horizontal position of the ball (as a function of time)?

## Homework Equations

(delta)y= Vyt + .5agt^2
(delta)x= Vxt
Vyi= Vsin(theta)
Vxi=Vcos(theta)

## The Attempt at a Solution

On the drawing the diagram part I am very confused. I don't know where to begin drawing, if have go by the given hints!

For part B: (the vertical position of the ball as function of time.)
y(t)= Vsin(theta)- gt^2/2
I do not think this is right because the problem specifies that the function should only be in terms of theta and v. I do not know how to get this function without g!

For part C: i got the horizontal position of the ball to ne:
x(t)= Vcos(theta)t

Related Introductory Physics Homework Help News on Phys.org
tiny-tim
Homework Helper
For part B: (the vertical position of the ball as function of time.)
y(t)= Vsin(theta)- gt^2/2
I do not think this is right because the problem specifies that the function should only be in terms of theta and v. I do not know how to get this function without g!

For part C: i got the horizontal position of the ball to ne:
x(t)= Vcos(theta)t

Hi ScullyX51! (have a theta: θ and a squared: ² )

You are always allowed to use g ! Then find out which value of θ grazes the wall. 