Projectile motion on inclined plane

[songoku]

Homework Statement

As shown in the figure below, small object A is projected upward along a smooth slope forming angle α with the horizontal plane, with initial speed v₀ in the direction of angle θ from the horizontal within the slope’s plane. The object continues to travel along the surface of the slope. Let us denote as h the maximum height that A attains on the slope, where height is measured vertically from A’s starting point. Find v₀

Homework Equations

projectile motion

The Attempt at a Solution

I interpret the question as projectile motion with total angle (α + θ) with respect to horizontal plane.

At maximum height, vertical speed is zero, so:
v_y² = v_oy² - 2gh

v₀ = √(2gh) / (sin (α + θ))

But the answer is v₀ = √(2gh) / (sin θ)

Do I interpret the question wrongly? Or there is mistake in my working?

 

[CWatters]

The angles don't add in the way you think.

 

[arpon]

I interpret the question as projectile motion with total angle (α + θ) with respect to horizontal plane.

You misinterpreted the question. The ball moves "within" the slopes plane, I mean, it does not go "above" the plane.

 

[songoku]

The angles don't add in the way you think.

You misinterpreted the question. The ball moves "within" the slopes plane, I mean, it does not go "above" the plane.

I get the hint. Thanks a lot