# Maximum perpendicular height above an inclined plane

1. Jul 31, 2013

### Woolyabyss

1. The problem statement, all variables and given/known data
A particle is projected from a point o with initial velocity u up a plane inclined at an angle 60° to the horizontal. The direction of projection makes an angle ∅ with the horizontal. The maximum perpendicular height is H.

prove that H = u^2( sin^2(∅) ) / g

2. Relevant equations

v = u + at

s = ut + .5(a)(t^2)

3. The attempt at a solution

the horizontal axis ( i axis ) is the line of greatest slope of the inclined plane. The j axis is perpendicular to the i axis.

splitting the u vector into horizontal and vertical components along the i and j axis'.

i axis.............. ucos( ∅ - 60 )

j axis..............usin( ∅ - 60)

force of gravity split into component vectors along and perpendicular to the inclined plane

i axis = gsin(60) = √3/2

j axis = (1/2)g

the object will have reached its maximum perpendicular height when its vertical velocity(Vy) is 0

Vy = usin(∅ - 60) - (1/2)gt = 0

t = 2u( sin( ∅ - 60 ) )/g

The maximum perpendicular height Sy = H

Sy = ut + (1/2)at^2

Sy = usin(∅-60)(2u)(sin(∅-60)/g -(1/2)(1/2)g(4u^2)( sin^2(∅-60)/g^2 )

simplify

Sy = u^2( sin^2( ∅ - 60 ) ) /g

when I simplify using

sin(A-B)=sin A cos B - cos A sin B

It just makes the equation more awkward

I think I was right using ∅ - 60 since both angles are taken from the horizontal so you have to subtract to get the angle between the objects line of projection and the plane.

Any help would be appreciated.

2. Jul 31, 2013

### PeterO

I worry that there is an H value that results if ∅ = 30o, but it is not possible to project the particle up a 60o incline if you project it at 30o to the horizontal ???

Perhaps the condition ∅ > 60o is implied.

The expression H = u^2( sin^2(∅) ) / g is merely the maximum height reached by any projectile fired at an angle ∅ with the horizontal.

3. Jul 31, 2013

### barryj

Can someone explain the problem here???

4. Jul 31, 2013

### haruspex

That doesn't bother me. The equation merely gives a parabolic trajectory which includes the point of origin of the particle. The slope must either be a tangent to the trajectory or cut it in two places, with a finite section of the trajectory above the slope.
The given result is clearly wrong, though. If the particle is fired parallel to the slope then H must be 0. Woolyabyss' answer, u2 sin2(∅-60)/g, is much more reasonable. Maybe the question should have said it makes an angle ∅ with the plane.

5. Aug 1, 2013

### Woolyabyss

I'm not sure then, I think I'll just leave this question and move on.

Last edited: Aug 1, 2013
6. Aug 1, 2013

### ehild

You can throw that particle from the slope at an angle θ>60° to the horizontal. The maximum height it reaches depends on the vertical component of its velocity , independent of the slope. H=(usinθ)2/2g.

ehild

7. Aug 1, 2013

### haruspex

Yes, but it's asking for the "perpendicular height", which Woolyabyss and I interpret as the max distance from the plane.

8. Aug 1, 2013

### ehild

Maybe it should have written that θ is the angle with respect to the slope. Choosing y perpendicular to the slope, y=usin(θ)t-gcos(60) t2/2. As cos60=1/2, y=usin(θ)t-g t2/4. Its maximum is H=u2sin2(θ)/g, the given solution.

ehild