STEP I 1994 question 9 solution - The Student Room
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STEP I 1994 question 9 solution

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STEP I question 9

Let (x,y) be a point on the cannon-ball's trajectory, then:

x=ut\cos\alpha

y=ut\sin\alpha - \frac{1}{2}gt^2


Now if the distance from the origin is \sqrt{x^2 + h^2}, then we have the point (x,h), so writing the x and y coordinates with the trig function the subject:

\cos\alpha = \frac{x}{ut}

\sin\alpha = \frac{h + \frac{1}{2}gt^2}{ut}

But \cos^2\alpha + \sin^2\alpha = 1, so:

x^2 + h^2 + hgt^2 + \frac{1}{4}g^2t^4 = u^2t^2

\frac{1}{4}g^2t^4 - (u^2-gh)t^2 + h^2 + x^2 = 0, as required.


Since this equation arose from considering a point we assumed to be on the trajectory, then there has to exist a real value of t at which the cannon-ball is at this point. For a real value of t, we need a real value of t^2, so the solutions to the quadratic must be real, so using the discriminant:

u^4 - 2ghu^2 + g^2h^2 - g^2(h^2+x^2) \geq 0

u^2(u^2-2gh) \geq g^2x^2

x \leq \frac{u\sqrt{u^2 -2gh}}{g}, as required.


Okay, now we need to show that there is an angle of firing such that we have  x= \frac{u\sqrt{u^2 - 2gh}}{g} when y=h, so going back to the equations for x and y we get:

x=ut\cos\alpha \Rightarrow t=\frac{\sqrt{u^2 - 2gh}}{g\cos\alpha}

and h=ut\sin\alpha - \frac{1}{2}gt^2

substitute in the time from the x into the y:

\frac{g(u^2-2gh)}{2g^2} \sec^2\alpha - \frac{u\sqrt{u^2-2gh}}{g}\tan\alpha + h=0

(u^2-2gh)\tan^2\alpha - 2u\sqrt{u^2-2gh}\tan\alpha + u^2 = 0

For there to exist an angle of firing as described, we again need real roots to this quadratic, 4u^2(u^2-2gh) - 4u^2(u^2-2gh)\geq 0, which is of course true (and it really is an angle of firing).


Solution by ad absurdum

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