Projectile Motion and linear motion problem

[Clipse]

A projectile is launched at an angle α from the edge of a cliff of height H above sea level. If it falls
into the sea a distance D from the base of the cliff, prove that its maximum height above sea level is

H + [(Dtan α)^2]/4(H + Dtan α)

Newtons equations of linear motion in the i/j plane.


Well, I understand that the max height is at a height H + (something) with that something being Sjmax, when Vj=0.

Ive got my Sjmax = (usin α)^2/ 2g

This is where i get a little stuck... I've tried a couple of things but the fact that its a cliff is making my reasoning wrong, if it wasnt a cliff, you could also say that the max vertical height is obtained when the horizontal displacement is D/2, the christmas break has made the old mind a little blunt... Any hints? Thanks so much in advance.

 

[Rainbow Child]

You have to use the distance D, to find u.

In what axis is D, and when the projectile reaches the sea?

 

[Clipse]

Ya, I can see you need to find an equation for u, because that's the unknown that isn't in the answer.

D is in the i plane, so do I find equations for when Sj= -H, and Si=D?

 

[Rainbow Child]

Yes! That's the way!

 

[magesha]

Not the most elegant solution but here is a way:

lets split the trajectory into two: one part where it reaches the same height as the starting point, and second part where it goes into the sea. Let's use H₁, D₁ and H₂, D₂ to denote the heights and distances of each segment:

H₂ = H
D₁ + D₂ = D

The solution we are looking for is the height of the first parabola, H₁. Let's use u to denote the initial velocity of projection. By solving the equations of motion in the i and j planes, we can get the following:

D₁ = 2u² sin$$\alpha$$ cos$$\alpha$$ / g

H₁ = u²sin² $$\alpha$$/2g

for the second projectile, let's calculate the time of travel using the i direction:

D₂ = D-D₁ = ucos$$\alpha$$ * t

t = (D-D₁)/ucos $$\alpha$$ =D/ucos$$\alpha$$ - 2usin$$\alpha$$/g = Dsec$$\alpha$$/u - 2usin$$\alpha$$/g

use this in the equation for j direction; note here the initial velocity will be usin$$\alpha$$ pointing downwards:

H = usin$$\alpha$$ *t + 1/2*g*t² = Dtan$$\alpha$$ - 2u²sin²$$\alpha$$/g + g/2 * [ D²sec²$$\alpha$$/u² +4u²sin²$$\alpha$$/g² - 4Dtan$$\alpha$$/g]
= Dtan$$\alpha$$ - 2u²sin²$$\alpha$$/g + gD²sec²$$\alpha$$/2u² +2u²sin²$$\alpha$$/g - 2Dtan$$\alpha$$
= gD²sec²$$\alpha$$/2u² - Dtan$$\alpha$$
Rearranging terms:
so, u² = gD²sec²$$\alpha$$/[2(H+Dtan$$\alpha$$)]

Substituting:

H₁ = u²sin²$$\alpha$$/2g = gD²sec²$$\alpha$$ * sin²$$\alpha$$ / [2g *2(H + Dtan$$\alpha$$)]
= D²tan²$$\alpha$$/[4(H + Dtan$$\alpha$$)] // sin $$\alpha$$ * sec $$\alpha$$ = sin$$\alpha$$ / cos$$\alpha$$ = tan$$\alpha$$

total height = H₁ + H₂ = H + (Dtan$$\alpha$$) ² / 4 (H + Dtan$$\alpha$$) <------ solved!

I tried my best to maintain clarity but let me know if something is not clear!