Projectile Motion & Theta: Finding Tan (Theta) Using Equations of Motion

[DCC01]

Homework Statement

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

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

Homework Equations

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 :)

 

[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.

 

[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))

Add (1a) and (2):

(3) -3h=(-2d)(tan(theta)) as required.

LDO :p