Maximum perpendicular height above an inclined plane

[Woolyabyss]

Homework Statement

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

Homework Equations

v = u + at

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

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.

 

[PeterO]

Homework Statement

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

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

Perhaps the condition ∅ > 60^o 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.

 

[barryj]

Can someone explain the problem here?

 

[haruspex]

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

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.

 

[Woolyabyss]

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

 

[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

 

[haruspex]

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

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

 

[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