- #1

BeyondBelief96

- 15

- 2

## Homework Statement

A particle A moves along the line y = d (30 m) with a constant velocity

## Homework Equations

Kinematic Equations

## The Attempt at a Solution

So I listed out all knowns and unknowns for both particle A & B as follows:

Particle A: Xi = 0, Xf = ?, Vix = 3 m/s, Vif = 3m/s, a = 0 t = ? and y = 30

Particle B: Xi = 0, Xf = ?, Vix = 0 m/s, Vif = ?, ax = asin(theta), t = ?

Yi = 0, Yf = 30, Viy = 0 m/s, Vif = ?, ay = acos(theta), t = ?

From there I used the kinematic equation d = vit + 1/2at^2 for particle A and B and got the following:

Particle A: x = 3t, y =30

Particle B: x = 1/2asin(theta)t^2, y = 1/2acos(theta)t^2

I set each of the equations equal to each other and tried to solve the system of equations for theta as follows:

3t = 1/2asin(theta)t^2 (Eq1)

30 = 1/2acos(theta)t^2 (Eq2)

**3 = 1/2asin(theta)t so t = 6/asin(theta)**

**[/B]**

So I plugged t into Eq2:

30 = 1/2acos(theta)[6/asin(theta)]^2

from here im at a loss on how to solve for theta. To be honest, my algebra solving skills arent super spectacular, which is a bit sad haha. I was able to simplify this as:

30 = 1/2acos(theta)[36/a^2sin^2(theta)] == 5/3 = cot(theta)/asin(theta) .....aaaaaand this is where im lost.

So I plugged t into Eq2:

30 = 1/2acos(theta)[6/asin(theta)]^2

from here im at a loss on how to solve for theta. To be honest, my algebra solving skills arent super spectacular, which is a bit sad haha. I was able to simplify this as:

30 = 1/2acos(theta)[36/a^2sin^2(theta)] == 5/3 = cot(theta)/asin(theta) .....aaaaaand this is where im lost.