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

2. ## Re: projectile

Originally Posted by markosheehan
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=[FONT="] √(2h/g)
i have no idea idea about the equation [/FONT]
How did you get that? That is what you should get if u= 0.

3. ## Re: projectile

the initial speed in the y direction is 0 i take it no?

"the trajectory being out to sea in a plane perpendicular to the cliff." does this not mean its projected horizontally?

ive tried putting the formula given in the question into the -b formula but i do not get any where with it.

4. ## Re: projectile

The provided answer, based on the Q. as phrased, is simply wrong!

Consider the vertical motion. Initially it is zero, assuming that the initial trajectory was horizontal. (We are not told otherwise!) The particle descend a distance h under the acceleration of gravity, g.
Using s = ut + ˝gt˛, we get h = 0 + ˝gt˛, from which t = √(2h/g).

The main problem is the wording of the Q.: it is vague; possibly incomplete. There is no place in mathematics for vagueness.

Al.