# A projectile fired at a speed of u and an angle of x to the horizontal (1 Viewer)

#### Karldahemster

##### Student
A projectile fired at a speed of u and an angle of x to the horizontal, reaches a
maximum height of H. When the same projectile is fired again at the same
angle with a new higher initial speed it reaches a new maximum height of 2H.

What is the new initial velocity in terms of u.

#### tazhossain99

##### Member
The new initial velocity should be 'root 2 times u'. Tell me if there's something wrong with it or you don't understand a part of it: https://imgur.com/PuPIDE5

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#### Powereaper

##### New Member
That's for the new uy. But for the new initial velocity, which is root of uy2 + ux2, I got u times root (sin squared x plus 1).
Not sure if it's correct, so a confirmation would be appreciated.

#### InteGrand

##### Well-Known Member
That's for the new uy. But for the new initial velocity, which is root of uy2 + ux2, I got u times root (sin squared x plus 1).
Not sure if it's correct, so a confirmation would be appreciated.
$\bg_white \noindent If the angle of projection is \theta and the speed of projection is u, then y = ( u\sin \theta)t - \frac{1}{2}gt^2. The maximum height thus occurs when t = \frac{u\sin \theta}{g}.$

$\bg_white \noindent Therefore, the maximum height is$

\bg_white \begin{align*}H &= (u\sin \theta)\times \frac{u\sin \theta}{g} -\frac{1}{2}g \times \frac{u^2 \sin^2 \theta}{g^2} \\ &= \frac{u^2 \sin^{2} \theta}{2g}.\end{align*}

$\bg_white \noindent So as we can see, H \propto u^2, that is, u\propto \sqrt{H}. Therefore, to double the maximum height, we must multiply the initial speed u by \sqrt{2} (for fixed \theta). So the answer is \boxed{\sqrt{2}u}.$

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#### InteGrand

##### Well-Known Member
That's for the new uy. But for the new initial velocity, which is root of uy2 + ux2, I got u times root (sin squared x plus 1).
Not sure if it's correct, so a confirmation would be appreciated.
$\bg_white \noindent Oh, I realised tazhossain99 answered the question above. And his working out is actually showing it for the new u, not new u_y.$

#### Powereaper

##### New Member
Oh,got it now. I thought horizontal velocity was the same, until I realised the question didn't mention that. Thanks for the help.