Projectile Motion Ball Calculation

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richievuong
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A question from my pre-unit test review

A ball is projected from point A with an intial velocity Vo, which is perpendicular to the incline shown. Knowing that the ball strikes the incline at B, determine the range R in terms of Vo and β.

Diagram can be seen here:

http://img146.imageshack.us/img146/3231/projectile001xu5.jpg
I drew the delta X/Y, and 90-βI did some calculations(very messy), still confused about it though

For typing purposes I used V1 for Vo
V1y = V1sin(90-β)
V1x = V1cos(90-β)

First I tried to find time:
y = V1yt + 1/2ayt²
0 = V1sin(90-β)t + 1/2(-9.8)t²
V1sin(90-β)t = 4.9t²
t = V1sin(90-β) / 4.9

Horizontal range:
x = V1xt
x = [V1cos(90-β)] [V1sin(90-β) / 4.9]

Finding R:
cosβ = X / R
cosβ = [V1cos(90-β)] [V1sin(90-β) / 4.9] / R
cosβ = [V1cos(90-β)V1sin(90-β) / 4.9R]

This looks really messed up can someone check my work please...if its too messy to read I'll write it out on request and scan it
 
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richievuong said:
First I tried to find time:
y = V1yt + 1/2ayt²
0 = V1sin(90-β)t + 1/2(-9.8)t²
V1sin(90-β)t = 4.9t²
t = V1sin(90-β) / 4.9
You are finding the time when y = 0. Why?

Horizontal range:
x = V1xt
x = [V1cos(90-β)] [V1sin(90-β) / 4.9]
This seems to be the x-coordinate when y = 0. Not what is needed.


Here's what I would suggest: Express y and x as functions of time. Combine this with an equation which relates y & x along the incline. (What's the equation of a straight line?)