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

*g*^{2}*t*^{4 }– 4(*u*^{2} + *gh*)*t*^{2} + 4(*h*^{2} + *d*^{2}) = 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]