a STEP-like mechanics question

Another equivalent ones is $\displaystyle \theta = \arctan \left(\frac v{\sqrt{v^2+2gh}} \right)$

Can I check out how you solved the equation when you set dR/dtheta = 0?

From $\displaystyle y = h + x\tan(\theta) - \frac{g}{2u^2}x^2(1+\tan^2 ( \theta ) ),$

We get

$\displaystyle \frac{dx}{d\theta} = \frac{x \sec^2(\theta)[\frac{x}{L}\tan(\theta)-1]}{ \tan(\theta)-\frac{x}{ L }(1+\tan^2(\theta)) }$

Where $L = \frac{u^2}{g}$

=0 when $\tan \theta = \frac{L}{x}$, can you see where to go from here?

Yeah, this was my way, I didn't know you combined the horizontal and vertical displacements to give a cartesian form of displacement and implicitly differentiated. I thought you differentiated R wrt theta which would've given loads of ^-1/2 terms which made me wonder how you solved that trig equation lol, fair enough

y= u sin theta t + 1/2at^2

x= u cos theta * t

t = x / (u cos theta)

Sub t into y, cartesian form boom.

-Write down the angle needed to project an object on a flat surface so that it achieves maximum horizontal displacement. Ignore air resistance.

-A projectile is projected with speed v at angle theta to the horizontal. The projectile is on a flat platform raised h metres above the ground, and the ground is flat as well. Derive an expression for theta which gives the maximum horizontal displacement of the projectile. Ignore air resistance.


http://postimg.org/image/o23uzglcx/

Can you PM me the answers please mate

Where are you stuck and what have you done so far?