STEP I 1994 question 9 solution

Register

About Us | Help | Sign in

Create New Article | Editing Help

Advanced Search

Edit \| Hide
Welcome \| Study Help \| A Levels \| Applications, UCAS etc. \| University Discussion \| Careers \| General Discussion \| Have Your Say \| Health & Relationships \| Debate & Discussion \| Technology \| Sport \| Entertainment \| Chat \| Ask A Moderator

STEP I 1994 question 9 solution

From The Student Room

Jump to:navigation, search

TSR Wiki > Study Help > Subjects and Revision > Mathematics > STEP > STEP I 1994 question 9 solution

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).