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How would I solve part a and b.

You can use double angle identities. Specifically, \cos(2\alpha)=\cos^2(\alpha)-\sin^2(\alpha) and \sin(2\alpha)=2\cos(\alpha)\sin(\alpha)

When you use the equation s_x=\frac{u^2\sin(\theta)}{g}, you have to use the initial velocity, but the question only gives the initial horizontal velocity. I think that's where you went wrong.

Maybe try and solve the square then expand it \\(1+2x-x^2)^5\\\\=(-(x^2-2x-1))^5\\\\=(-((x-1)^2-2))^5\\\\=(2-(x-1)^2)^5 then expand it from here

This is more like a cosmetic tip but if you want the brackets to cover the entire fraction use "\left" and "\right", for example "\left(\frac{2\pi}{9}\right)" results in \left(\frac{2\pi}{9}\right). So yeah, use it if you want to.

I have heard about the auxiliary method but I haven't learnt it. Also, thankyou! Not really sure, this was just part of the homework given to me by my advanced teacher. I probably will ask her were she got this.

How would is solve a question like this.

Cambridge Extension 2 Maths 2020

Thankyou!

Q10 part d and Q11 part d, I am kinda confused as to what to do.

I think another good related text would be 'Night' by Elie Wiesel, which is a memoir (around 100 pages) about the experiences of the author during WW2 surviving in Nazi concentration camps with his father. It's a sad and depressing story, but I think it would closely relate to the human...

Thankyou! Now it makes much more sense.

I need help with part c thankyou!

You could assume that the point K extends until it is directly below point B, forming the right angle triangle BKH. Since we already have the length of BH from part a. We can find the length of BK by subtracting the heights of the balloon and the helicopter. \\\text{Balloon...

u_y=0 when the projectile is launched horizontally. You can find s_y through the following methods. 1. Use s_y=u_yt+\frac{1}{2}(-g)t^2, since u_y=0 we can simplify the expression to s_y=-\frac{1}{2}gt^2 2. Use v_y^2=u_y^2+2(-g)s_y, which can be rearranged to s_y=-\frac{v_y^2}{2g}

Edited my above post.

For part b \\y=\tan(2\theta)\\y=\frac{2\tan(\theta)}{1-\tan^2(\theta)}\\\\\tan(\theta)=x-1\\\\y=\frac{2(x-1)}{1-(x-1)^2}\\\\y=\frac{2x-2}{-x^2+2x}

But anyway, good luck to the people who are doing English Ext 2! 😀