Projectile motion shortcuts: is H = R tanθ / 4 legit? and why do 30° and 60° give the same range?

11 answers & replies

• [+][–]Arjun Nair12thSample· 10d ago

  Derivation of the magic one: divide $H$ by $R$.

  $\frac{H}{R} = \frac{u^2\sin^2\theta/2g}{u^2\cdot 2\sin\theta\cos\theta/g} = \frac{\sin\theta}{4\cos\theta} = \frac{\tan\theta}{4}$

  So $H = \frac{R\tan\theta}{4}$, or $\tan\theta = \frac{4H}{R}$. Your $R = 4H$ gives $\tan\theta = 1$, so 45°. Correct. The formula holds only for level ground (same launch and landing height).

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  • [+][–]Rohan Sharma11thSample· 10d agoAuthor

    oh that's just division of two formulas, I'm dumb 💀 thanks

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• [+][–]Kabir Joshi11thSample· 10d ago

  For the 30°/60° question: range is the same, max height is not. Heights are $H_1 = \frac{u^2\sin^2 30°}{2g}$ and $H_2 = \frac{u^2\sin^2 60°}{2g}$. Ratio is $\frac{\sin^2 30°}{\sin^2 60°} = \frac{1/4}{3/4} = 1:3$. In general for complementary angles $\frac{H_1}{H_2} = \tan^2\theta$.

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• [+][–]Harsh Gupta12thSample· 10d ago

  the 60° throw goes higher and takes longer but lands at the same spot. basically the 60° is dramatic and the 30° is efficient 😌

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• [+][–]Aman YadavDropperSample· 10d ago

  Time of flight also differs. For θ and 90°−θ: $T_1 = \frac{2u\sin\theta}{g}$, $T_2 = \frac{2u\cos\theta}{g}$, so $\frac{T_1}{T_2} = \tan\theta$. And $T_1 T_2 = \frac{2R}{g}$, which shows up in some questions. Only the horizontal range is shared.

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• [+][–]Ishaan Bose11thSample· 10d ago

  wait is max height also same at 45° since sin²45 = 1/2? or no

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  • [+][–]Tushar PawarDropperSample· 10d ago

    At 45° the max height is $\frac{u^2}{4g}$ and the range is $\frac{u^2}{g}$, so $R = 4H$, which is your tanθ = 1 case again. Nothing special about it being "same", it just gives the neat ratio.

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• [+][–]Vivek Mishra12thSample· 10d ago

  For the cliff question: the formulas for T, H, R above are for same-level. For a cliff, you split in x and y. Horizontal: $x = u\cos\theta\cdot t$. Vertical: $y = h + u\sin\theta\cdot t - \frac{1}{2}gt^2$. Put $y = 0$, solve the quadratic for t, then find x. I know it takes more time but it is safe and it works for all cases. The shortcut formulas are only the special case when h = 0.

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  • [+][–]Rohan Sharma11thSample· 10d agoAuthor

    this is the thing I needed. so basically shortcuts = derived from the full equations for one special case. ok. I'll derive each one once and then use them

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• [+][–]Neel Chatterjee12thSample· 9d ago

  Good plan. Also a quick check for the case where it's thrown horizontally from height $h$ (θ = 0): $t = \sqrt{2h/g}$ and $R = u\sqrt{2h/g}$. Easy to get wrong if you apply the level-ground $T$ formula, which would give 0.

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  • [+][–]Rohan Sharma11thSample· 9d agoAuthor

    oh no that's exactly what I did in a quiz last week lol. thanks Neel

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