# Maximizing range of projectile on a wedge

1. Apr 13, 2015

### AlephNumbers

1. The problem statement, all variables and given/known data
A greasy spud is launched off of the top of an infinitely large frictionless wedge. The angle that the wedge slopes down at is 30 degrees. Find the launch angle that maximizes the horizontal range of the spud.

2. Relevant equations
x = x0 + vxt + (1/2)at2

3. The attempt at a solution
I began by rotating the axes so that the x-axis would be parallel with the wedge. Then I divided gravity into x and y components.

ax = sin(30)mg

ay = cos(30)mg

Then I found the total time the spud is in the air

t = 2(sinθv0)/(cos(30)mg)

The horizontal range of the spud can be written as

x = cosθv0t + (1/2)(sinθmg)t2

Substituting for t yields

x = ((2sinθv02)/(cos(30))(cosθ + (sin2θ)/(cos(30))

I then took the derivative of cos(θ) + sin2(θ))/(cos(30) with respect to θ and got

2cosθsinθ/cos(30) - sinθ

setting this equal to zero and solving for theta yielded 64.3 degrees. Then, subtracting the angle through which I rotated the axes gives 34.3 degrees.

Last edited: Apr 14, 2015
2. Apr 14, 2015

### ehild

Check the parentheses!
You made the job difficult with rotating the axes. The trajectory of the projectile, y(x) (height-horizontal displacement) does not depend on the slope, and the x range can be found where y(x) intersects the line of the slope.

3. Apr 14, 2015

### haruspex

Having rotated the axes, x no longer represents horizontal range.

4. Apr 14, 2015

### AlephNumbers

Right. That was a bit silly of me. So what I solved for was the angle that maximizes the distance covered on the wedge. For some reason I thought that if I maximized the distance that the spud would travel on the wedge I would also maximize the horizontal range.