Projectile Motion - need help

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SUMMARY

The discussion focuses on calculating the time required for a projectile, fired from the origin with an initial velocity \( v \) at an angle \( \alpha \), to cross a line defined by the equation \( y = ax \), where \( a \) is determined by the angle \( \beta \). The projectile's vertical position is described by the equation \( y = y_0 + v_0(\sin \alpha)t - 4.9t^2 \). To find the intersection of the projectile's path and the line, the user must express \( y \) in terms of \( \beta \), set it equal to the projectile's equation, and solve for \( t \) by determining the value of \( x \) at which both equations intersect.

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don_anon25
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Projectile Motion -- need urgent help :)

If a projectile is fired from the origin of the coordinate system with an initial velocity v and in a direction making an angle alpha with the horizontal, calculate the time required for the projectile to cross a line passing thorugh the origin and making an angle beta less than alpha with the horizontal.

I know that the position of the projectile is described by
y=y0+v0(sin alpha)*t-4.9t^2.

I let the line be y=ax. I also drew a triangle with Beta as the angle, y=ax as the opposite side, and x as the adjacent side. I'm assuming I want to find y in terms of Beta and then set this equal to y in my projectile motion equation...then solve for t?

Thanks in advance!
 
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don_anon25 said:
I know that the position of the projectile is described by
y=y0+v0(sin alpha)*t-4.9t^2.
Combine this with the equation for the horizontal position and you can get y as a function of x.

I let the line be y=ax. I also drew a triangle with Beta as the angle, y=ax as the opposite side, and x as the adjacent side. I'm assuming I want to find y in terms of Beta and then set this equal to y in my projectile motion equation...then solve for t?
Write the equation of the straight line. (What is "a" in terms of beta?)

Now you can find the value of x where the two functions intersect. And then plug that into one of the other equations to solve for the time.
 

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