May 14th, 2017, 11:37 AM  #1 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1  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 
May 14th, 2017, 01:26 PM  #2 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,600 Thanks: 816 
The trick for (i) is to solve for $\cos(\theta)$ from the xdirection equation of motion, and for $\sin(\theta)$ from the ydirection 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. 
May 14th, 2017, 01:36 PM  #3 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,600 Thanks: 816 
for (ii) just note that the maximum range is obtained at $\theta=\dfrac{\pi}{4}$ as usual and slog through the algebra.

May 15th, 2017, 12:08 AM  #4 
Member Joined: May 2016 From: Ireland Posts: 96 Thanks: 1 
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 
May 15th, 2017, 12:30 AM  #5 
Senior Member Joined: Sep 2015 From: Southern California, USA Posts: 1,600 Thanks: 816 
$x(t)=u \cos(\theta) t$ $y(t) = h + u \sin(\theta) t  \dfrac{ g t^2}{2}$ at the time of hitting the water. $d = u \cos(\theta)t$ $0 = h + u \sin(\theta)t  \dfrac{ g t^2}{2}$ $\cos(\theta) = \dfrac{d}{u t}$ $\sin(\theta) = \dfrac{g t^2  2h}{2}$ $1 = \left(\dfrac{d}{u t}\right)^2 + \left( \dfrac{g t^2  2h}{2}\right)^2$ $0 = \left(\dfrac{d}{u t}\right)^2 + \left( \dfrac{g t^2  2h}{2}\right)^21$ 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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