# Projectile motion and theta

1. Aug 25, 2009

### DCC01

1. The problem statement, all variables and given/known data

A projectile is launched from O with speed u at angle theta to the horizontal. In its subsequent trajectory it just clears two obstacles of height h at horizontal distances d and 2d from O, respectively.

http://img403.imageshack.us/img403/4369/projectile.jpg [Broken]

Show that tan (theta) = 3h/2d

2. Relevant equations

3. The attempt at a solution

I do not see anyway to get tan (theta) here. From class I have the following equations:

Solution of the equation of motion:
r(t)=(utcos(theta))i + (utsin(theta) - ½gt²)j

trajectory is parabola with equation:

y=xtan(theta)-x²(g/2u²)(1+tan²(theta))

but I cannot see how I begin to find just tan(theta)

I just need to know where to start. My guess is I'm missing something that's taken to be 'obvious' but what my professor thinks is 'obvious' is usually obscure for me :)

Last edited by a moderator: May 4, 2017
2. Aug 25, 2009

### kuruman

You need to relate theta to d and h. You also have an equation for the parabola y(x). Start by saying that y is equal to h at x = d and x = 2d. That's the condition for "just clearing" the obstacles.

3. Aug 25, 2009

### DCC01

So I get a pair of simultaneous equations for h:

(1) h=(d)tan(theta)-(d)²(g/2u²)(1+tan²(theta))
(2) h=(2d)tan(theta)-4(d)²(g/2u²)(1+tan²(theta))

Multiply (1) by -4:

(1a) -4h=(-4d)tan(theta)+4(d)²(g/2u²)(1+tan²(theta))