# Vertical / Horizontal Displacements

### Vertical / Horizontal Displacements

Hey guys! Firstly, let me start by apologizing if this is completely the wrong place to put this question. As you'll soon tell my maths is... well, questionable at best.

Let me just give some context to avoid any assumptions I'm starting an OU degree in autumn with my first module being mathematics. My maths knowledge didn't get any further than highschool so it's a bit of a challenge. I've been trying to revise using practise papers and bring myself at least up to scratch but this is proving difficult. There are some questions where I'm spending hours on them which would take some of you 10 minutes. But this question here is driving me nuts...

The question:

A boy standing on a level playing field kicks a ball into the air from the
ground at his feet. The trajectory of the ball can be modelled by the
equation

y=3x - x^2/2

where x is the horizontal displacement (in metres) of the ball from the
boy and y is the vertical displacement (in metres) measured from the
ground.

(i) Find the maximum vertical displacement of the ball.

(ii) If there is a 3 metre high wall at a distance of 5 metres from the
boy, in the direction the ball is travelling, will the ball be kicked

I have lots more questions like this but I want to work those out myself when I have an understanding of the first one :/

Sorry this is a pretty big 'text dump', but my math ability is pretty bad and I don't even know where to begin.

Cheers guys!
Jay
Guest

### Re: Vertical / Horizontal Displacements

(i) Find the maximum vertical displacement of the ball.

You're looking for the vertex of the given parabola. There's a few ways to do this and which one you choose is mostly a matter of personal preference. One of them is to memorize that the axis of symmetry of a parabola is always given by $$x=\frac{-b}{2a}$$. This formula applies to parabolas written as $$y = ax^2 + bx + c$$ so in your case, rewriting the equation as $$y = -\frac{1}{2} x^2 + 3x$$, $$a =1/2$$ and $$b = 3$$. This puts the maximum at $$x=\frac{-3}{2*-1/2} = 3$$. Now this x is the horizontal displacement at which the maximum height occurs, but the question asks for the max vertical displacement. So to finish it just find the value of y that corresponds to x=3 or $$y_{max } = 3*3 - \frac{1}{2} 3^2 = 9 - 9/2 =9/2.$$

Other possible approaches include completing the square and using the discriminant of the quadratic formula. If you want to see either of these worked out just ask.

(ii) If there is a 3 metre high wall at a distance of 5 metres from the
boy, in the direction the ball is travelling, will the ball be kicked

This is straightforward. You're given an equation for height in terms of distance, so just plug x=5 into that equation and find the corresponding height at that point. $$3*5 - 5^2/2 = 15-25/2 = 5/2$$. Bonk! The ball slams into the wall at a height of 2.5 metres.
Guest

### Re: Vertical / Horizontal Displacements

Ah! I see! Why was I overthinking it so much?! Thank you so much! You have no idea how long that question has been bugging me...

If I may be so bold... could I ask you to take a look at another question that has been driving me insane too? I've created a separate post: https://www.math10.com/forum/viewtopic.php?f=19&t=5851&p=11484#p11484

FYI: I'm the OP, never occurred to me to create an account originally!

Cheers
Jay

JJKD

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