How Does an Object's Trajectory Intersect with a Sloped Line?

[grinosaurus]

Hello happy online physics homework helpers! Long time listener, first time caller. Had a test today, and was stumped by a question about projectile motion - it's in the past now but I'd still like to figure it out. And - in the interest of full disclosure - I think we can do some corrections of out test to get a quarter of the marks missed.

Anyways, I'll be describing it from memory and it was originally drawn out... so hopefully this is coherent and correct.

Homework Statement

An object is launched into the air at initial velocity v, in a direction that is alpha degrees above a straight line (l). This line, in turn, is theta degrees above the horizontal. At what point does the object intersect with line l - your answer should be given in terms of v, alpha, theta and g.

Homework Equations

well,

vertical displacement = sin(α + θ)*t - g/2(t^2)
hortizontal displacement = cos(α + θ)*t

other than that, I'm really stumped.

The Attempt at a Solution

I know you guys are are sticklers for this, but I am quite bamboozled. Can I say that the slope of the straight line is tanθ, so it's equation is y=tanθ*x, then set equal to something?

Sheesh.

Thanks in advance!

 

[tedbradly]

Hello happy online physics homework helpers! Long time listener, first time caller. Had a test today, and was stumped by a question about projectile motion - it's in the past now but I'd still like to figure it out. And - in the interest of full disclosure - I think we can do some corrections of out test to get a quarter of the marks missed.

Anyways, I'll be describing it from memory and it was originally drawn out... so hopefully this is coherent and correct.

Homework Statement

An object is launched into the air at initial velocity v, in a direction that is alpha degrees above a straight line (l). This line, in turn, is theta degrees above the horizontal. At what point does the object intersect with line l - your answer should be given in terms of v, alpha, theta and g.

Homework Equations

well,

vertical displacement = sin(α + θ)*t - g/2(t^2)
hortizontal displacement = cos(α + θ)*t

other than that, I'm really stumped.

The Attempt at a Solution

I know you guys are are sticklers for this, but I am quite bamboozled. Can I say that the slope of the straight line is tanθ, so it's equation is y=tanθ*x, then set equal to something?

Sheesh.

Thanks in advance!

x_{v}(t) = sin(\alpha + \theta)vt - \frac{gt^2}{2}
x_{vL}(t) = x_{h}(t)tan(\theta)

where x(t) is a function of displacement relative to the tossing point, the subscript h means horizontal, the subscript v means vertical, and the subscript L means 'for the line' (as opposed to 'for the tossed object')

You are interested in finding when these two equations equal each other. Note, the height for the line is its slope (y/x) times how much x it has displaced, which is always equal to how much x the tossed object has displaced.

 

[ashishsinghal]

the best and most efficient way is to take x-axis along the given line. In this way you will have to resolve acceleration due to gravity in these new axes. Not much complication in this way I hope...

 

[ashishsinghal]

yeah...forgot something, here displacement in y is zero - no problem at all now huh? enjoy