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

Karldahemster · Dec 6, 2017

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 · Dec 6, 2017

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

Powereaper · Jan 8, 2018

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 · Jan 9, 2018

Powereaper said:
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.

$\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 $t = \frac{u\sin \theta}{g}$.$

$\noindent Therefore, the maximum height is$

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

$\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 $\boxed{\sqrt{2}u}$.$

InteGrand · Jan 9, 2018

Powereaper said:
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.

$\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 · Jan 9, 2018

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

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

Karldahemster

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InteGrand

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InteGrand

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