- #1
grinosaurus
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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.
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.
well,
vertical displacement = sin(α + θ)*t - g/2(t^2)
hortizontal displacement = cos(α + θ)*t
other than that, I'm really stumped.
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!
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!