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Determine the angle of projection of a projectile if its speed at maximum height is $\sqrt{\frac{2}{5}}$ of its speed at half the maximum height.

My solution:

$$H_{max}=\frac{{v_0}^2\sin^2(\theta)}{2g}\implies \frac{1}{2}H_{max}=\frac{{v_0}^2\sin^2(\theta)}{4g}\\v_{x}=v_0\cos(\theta)\quad {v_{\frac{H}{2}y}}^2={v_0}^2\sin^2(\theta)-2g\left(\frac{{v_0}^2\sin^2(\theta)}{4g}\right)=\frac{1}{2}{v_0}^2\sin^2(\theta)\\v_0\cos(\theta)=\sqrt{\frac{2}{5}}\sqrt{{v_0}^2\cos^2(\theta)+\frac{1}{2}{v_0}^2\sin^2(\theta)}\\\cos(\theta)=\sqrt{\frac{2}{5}}\sqrt{\cos^2(\theta)+\frac{1}{2}-\frac{1}{2}\cos^2(\theta)}\\\cos^2(\theta)=\frac{1}{5}\cos^2(\theta)+\frac{1}{5}\\\cos^2(\theta)=\frac{1}{4}\implies\cos(\theta)=\frac{1}{2}\implies\theta=\frac{\pi}{3}$$ Solution found on another website: $$gH_{max}=\frac{{v_0}^2\sin^2(\theta)}{2}\quad {v_x}^2={v_0}^2\cos^2(\theta)\\{v_{\frac{H}{2}}}^2={v_0}^2-2g\left(\frac{1}{2}H_{max}\right)={v_0}^2-\frac{{v_0}^2\sin^2(\theta)}{2}\\v_{0}\cos(\theta)=\sqrt{\frac{2}{5}}v_{\frac{H}{2}}\implies {v_{0}}^2\cos^2(\theta)=\frac{2}{5}{v_{\frac{H}{2}}}^2=\frac{2}{5}\left({v_0}^2-\frac{{v_0}^2\sin^2(\theta)}{2}\right)\\5\cos^2(\theta)=2-\sin^2(\theta)=1+\cos^2(\theta)\\\cos^2(\theta)=\frac{1}{4}\implies\cos(\theta)=\frac{1}{2}\implies\theta=\frac{\pi}{3}$$

What I don't quite understand in the second solution is the application of the kinematics formula $v^2={v_0}^2+2a\Delta d$ (second line). I thought the formula held only for one dimensional kinematics, but its usage here would imply two dimensional vector addition since the initial velocity and gravity aren't parallel vectors. Can someone help clarify this for me?

Anson Pang
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1 Answers1

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You know that

$$v_y^2 = v_{0y}^2 + 2a_y\Delta y$$ and that $$v_x^2 = v_{0x}^2 + 2a_x\Delta x.$$

First,

$$v^2 = \textbf{v}\cdot\textbf{v} = \left(v_x \hat{\textbf{x}} + v_y \hat{\textbf{y}}\right)\cdot\left(v_x \hat{\textbf{x}} + v_y \hat{\textbf{y}}\right) = v_x^2 + v_y^2.$$

Second,

$$v_0^2 = \textbf{v}_0\cdot\textbf{v}_0 = v_{0x}^2 + v_{0y}^2,$$

and

$$2a_x\Delta x = 0.$$

Last, $$v^2 = v_0^2 - 2g\Delta y.$$

  • By the way, despite being a fourth-year physics major, I didn't know this until now. I just figured it out. I learned something thanks to your question. –  Jan 12 '19 at 05:19
  • If you don't know dot products, you can probably still follow this if you ignore them. –  Jan 12 '19 at 05:21
  • Can you provide a bit of an explanation? – Anson Pang Jan 12 '19 at 06:19
  • You don't understand why $v^2 = v_x^2 + v_y^2$ is what you mean? –  Jan 12 '19 at 06:42
  • If so, then think of it geometrically: $v_x$ and $v_y$ are two legs of a right triangle and $v$ is the hypotenuse---Pythagorean theorem. –  Jan 12 '19 at 06:52
  • Oh wait got it. Thanks – Anson Pang Jan 12 '19 at 17:39