Maximum perpendicular height above an inclined plane

mercredi 31 juillet 2013

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.