Projectile Motion - Help (1 Viewer)

[the-derivative]

Hello guys,

I'm having trouble with this projectile motion question, wondering if you guys can help me (it's such a bitch).

A particle is projected upwards at an angle A with a speed V relative to the firing platform which is stationary and inclined at 30 degrees to the horizontal. The particle strikes the platform at right angle.

a. Using Cartesian coordinates, calculate the coordinates of the point P.
b. Show that the range of the inclined plane is 4V^2/7g.

I've managed to get the Cartesian equation of the projectile, but that's about it. If you guys can help, I'd really appreciate it.

Thank you in advance. =)

 

[gurmies]

Question is quite lengthy - provided there's no simpler way. I would solve the cartesian equation of the path of the particle simultaneously with the line represented by the inclined plane i.e. y = xsqrt3. This will yield an extremely ugly x co-ordinate which appears useless, however if you take into account that the particle lands on the plane at right angles, you can differentiate and take m1m2 = -1 to get a condition. This condition will simplify the original x value of the co-ordinate of P and from there you can find the y co-ordinate. Probably best off using distance formula here with the origin to work out the range of the plane...As I said, not exactly an aesthetic solution.

Sorry for large picture - also there's a mistake somewhere, I've written P = 6v^2/21g which is meant to be y.

 

[the-derivative]

Wow, oh my god!

Thank you so much. Today in class, me and my friend spent a while, but we couldn't really interpret the question properly. Again, thanks heaps.

So smart ^_^

Also, I got stuck at the coordinate and never thought of differentiating and using the fact they were right angles. Genius. haha.

 

[gurmies]

Hm, I was also stuck at that part, but I gave it a go and it gives a nice and utilisable condition =] For a while I thought "Is it even necessary to know it hits at right angles?"

 

[Timothy.Siu]

x=Vtcos@ y=-0.5gt^2 + Vtsin@

dx/dt=Vcos @ (assuming @ is a constant)
dy/dt=-gt+Vsin@
dy/dx=-gt/Vcos@ + tan @
dy/dx=-root3 when they intersect since gradient of the incline is 1/root3

equation of the platform is y=x/root3
putting g=10
subbing in x and y, -5t^2+Vtsin@=V.t cos@/root3
t(Vcos@/root3 + 5t-Vsin@)=0
t=/=0
t=(Vsin@-Vcos@/root3)/5
subbing t into -gt/Vcos@ + tan @=-root3
-gtan@/5+g/5root3 + tan@=-root3
tan@=5root3/3
@=arctan 5root3/3
i think

this means that angle is independent of velocity..
there is only one angle that the projectile can be fired at. (dont know if this is right then)