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May 14th, 2017, 10:37 AM   #1
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projectile

A particle is projected at initial speed u from the top of a cliff of height h, the trajectory being out to sea in a plane perpendicular to the cliff.
The particle strikes the sea at a distance d from the foot of the cliff.
(i) Show that the possible times of flight can be obtained from the equation

g2t4 – 4(u2 + gh)t2 + 4(h2 + d2) = 0.
(ii) Hence or otherwise, prove that the maximum value of d for a particular u and h is
im stuck on both parts.
i started this off by using s=ut+.5at^2 take the projected point as the origin -h=0(t)+.5(-g)(t)^2 as the initial speed in the y direction is zero as its projected horizontally and when it hits the bottom its traveled -h in the y direction. this gets me t= √(2h/g)
i have no idea idea about the equation
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May 14th, 2017, 12:26 PM   #2
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The trick for (i) is to solve for $\cos(\theta)$ from the x-direction equation of motion, and for $\sin(\theta)$ from the y-direction and then note that

$\cos^2(\theta)+\sin^2(\theta) = 1$

wave the magic algebra wand and your given equation pops out.

I'll look at (ii) while you're digesting this.
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May 14th, 2017, 12:36 PM   #3
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for (ii) just note that the maximum range is obtained at $\theta=\dfrac{\pi}{4}$ as usual and slog through the algebra.
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May 14th, 2017, 11:08 PM   #4
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I'm having trouble forming it.
So I got cos @=x/it in the x direction
And in the y direction sin@=(2y+gt^2)/2ut
And when I sub this into the cos sin formula I get
4x^2 + 4y^2 +g^2t^2 =4u^2t^2
I'm not sure where to go from here x and y stand for distances in the x and y direction.
I'm not sure what h2 u2 and d2 stand for
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May 14th, 2017, 11:30 PM   #5
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$x(t)=u \cos(\theta) t$

$y(t) = h + u \sin(\theta) t - \dfrac{ g t^2}{2}$

at the time of hitting the water.

$d = u \cos(\theta)t$

$0 = h + u \sin(\theta)t - \dfrac{ g t^2}{2}$

$\cos(\theta) = \dfrac{d}{u t}$

$\sin(\theta) = \dfrac{g t^2 - 2h}{2}$

$1 = \left(\dfrac{d}{u t}\right)^2 + \left( \dfrac{g t^2 - 2h}{2}\right)^2$

$0 = \left(\dfrac{d}{u t}\right)^2 + \left( \dfrac{g t^2 - 2h}{2}\right)^2-1$

If you multiply both sides by $4 u^2 t^2$ and collect things by $t$ you get

$g^2 t^4 -4 (u^2 +gh)t^2+4(d^2 + h^2) = 0$

This is the equation you are given.
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