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 a = h(c+d)/c d

   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
   

   ___ Ö2g h

   
   , 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

   x² = 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

   xtana- (x²)/4hcosa is less than h-(x²)/4h,

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

Q1: Under the usual notation:

(1) c=v t cosa

(2) h=v t sina-g t² /2

(3) c+d=v² sin2a/g

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

(4) h=c tana- (g c² /2v² ) sec² a

Eliminate v (from (3) and (4)) gives: (5) h=c tana- [c² /2(c+d)] sin 2asec² a = {c-[c²/(c+d)]} tana

which gives the required answer.

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

y=xtana-(x²/4h)sec²a = -x²/4h+xtana-(x²/4h)tan²a

So

x²tan² a- 4h x tana+ (4h y+x²) = 0

Condition for no real roots gives (2h x)² < x² (4h y+x²). 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)!