Projectile motion on an inclined plane

[songoku]

Homework Statement

An object move in trajectory as shown. Find:
a. the time when the particle falls to the plane
b. tan⁡ θ[SUB]i[/SUB] so that the particle hits the plane vertically

Relevant Equations

Projectile motion

a. I tried to "rotate" the inclined plane so the surface of the inclined plane becomes horizontal

h = V_i sin θ_i . t - 1/2 g cos ∅ t² and when it falls to the plane, y = 0 so:
0 = V_i sin θ_i . t - 1/2 g cos ∅ t²
t = (2 V_i sin θ_i) / (g cos ∅)

Is this correct?b. Particle hits the plane vertically means that vertical component of the velocity must be zero. Since the vertical component of velocity of the object will never be zero if we take horizontal line as reference so I think I need to take the surface of inclined plane as reference.

V_x = 0
V_i cos θ_i - g sin ∅ . t = 0

Then I don't know how to continue

Thanks

 

[haruspex]

Particle hits the plane vertically means that vertical component of the velocity must be zero.

No, if it hits vertically then the horizontal component is zero. But that will only happen if it was vertical on take off, so makes no sense.
I would think question should specify either horizontally or normal to the plane.

 

[Orodruin]

Vi cos θi - g sin ∅ . t = 0

Assuming the question means that the projectile hits the plane at a 90 degree angle to the plane: You already have an expression for $t$ for when the projectile hits. Insert this and solve for $\tan\theta_i$.

 

[songoku]

No, if it hits vertically then the horizontal component is zero. But that will only happen if it was vertical on take off, so makes no sense.
I would think question should specify either horizontally or normal to the plane.

Ah yes, my bad. I mean horizontal component of velocity is zero. I think horizontally is impossible so I assume the question is referring to normal to the plane.

Assuming the question means that the projectile hits the plane at a 90 degree angle to the plane: You already have an expression for $t$ for when the projectile hits. Insert this and solve for $\tan\theta_i$.

Oh ok, don't realize I can use time from question (a)

Thank you very much for the help haruspex and orodruin

 

[neilparker62]

I am not sure you can just "rotate the plane" to solve for t. The projectile is launched at an angle $(θ+∅)$ to the horizontal. Unless angle θ is supposed to be between projectile vector and horizontal rather that between projectile vector and inclined plane as appears to be the case in your diagram ?

 

[Orodruin]

I am not sure you can just "rotate the plane" to solve for t. The projectile is launched at an angle $(θ+∅)$ to the horizontal. Unless angle θ is supposed to be between projectile vector and horizontal rather that between projectile vector and inclined plane as appears to be the case in your diagram ?

The OP has done that part correctly.

 

[songoku]

I am not sure you can just "rotate the plane" to solve for t. The projectile is launched at an angle $(θ+∅)$ to the horizontal. Unless angle θ is supposed to be between projectile vector and horizontal rather that between projectile vector and inclined plane as appears to be the case in your diagram ?

I "rotate the plane" so that the object moves with initial angle θ_i and I also "rotate" the acceleration of free fall so it has component in both horizontal and vertical direction

 

[neilparker62]

Ok - sorry. Missed that you were also 'rotating' g.