Could someone please help me with the following two questions:

1. A particle is projected under the influence of gravity from a point O on a level plane in such a way that, when its horizontal distance from O is $c$, its height is $h$. It then lands on the plane at a distance $c+d$ from O. Show that the angle of projection satisfies

   $\tan \alpha =h(c+d)/cd$

   The second part of this question involves deriving the conditions that the speed of projection must satisfy, but if someone could help me with the first part then I should be able to sort this out.

2. A particle is projected from a point O with speed $\sqrt{2gh}$, where $g$ is the accn. due to gravity. Show that it is impossible, whatever the angle of projection for the particle to reach a point above the parabola

   $x^2=4h(h-y)$

   where $x$ is horizontal distance from O and $y$ is vertical.

   I have gotten to a point where I have to prove that

   $x\tan \alpha -(x^2)/4h\cos \alpha$ is less than $h-(x^2)/4h$,

   however, I can't prove this and may be doing the completely wrong thing!

Q1: Under the usual notation:

(1) $c=vt\cos \alpha$

(2) $h=vt\sin \alpha -g{t}^2/2$

(3) $c+d=v^2\sin 2\alpha /g$

Eliminate $t$ (from (1) and (2)) gives:

(4) $h=c\tan \alpha -(g{c}^2/2v^2){\sec }^2\alpha$

Eliminate $v$ (from (3) and (4)) gives: (5) $h=c\tan \alpha -[c^2/2(c+d)]\sin 2\alpha {\sec }^2\alpha =\{c-[c^2/(c+d)]\}\tan \alpha$

which gives the required answer.

For Q2, suppose the point $(x,y)$ is on the trajectory with projection angle $\alpha$. Then we obtain

$y=x\tan \alpha -(x^2/4h){\sec }^2\alpha =-x^2/4h+x\tan \alpha -(x^2/4h){\tan }^2\alpha$

So

$x^2{\tan }^2\alpha -4hx\tan \alpha +(4hy+x^2)=0$

Condition for no real roots gives $(2hx{)}^2<x^2(4hy+x^2)$. Rearranging gives the answer stated.

Kerwin

Thanks Kerwin, I didn't think about deriving (3) in the first question and forgot that the second one could be rearranged into a quadratic in tan alpha. Hopefully I'll have improved by the time I come to take my STEP exams (if I take them at all)!