Just a simple calculus question

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SUMMARY

The discussion centers on the mathematical derivation of the projectile motion formula, specifically for calculating the angle of elevation for maximum range when launching from a height \( h \). The equation \(\frac{v_0^2 \sin(2a)}{2g} + \frac{v_0 \cos(a)}{g} \sqrt{v_0^2 \sin^2(a) + hg}\) is shown to be equivalent to \(\frac{v_0^2 \sin(2a)}{2g} (1 + \sqrt{1 + \frac{2hg}{v_0^2 \sin^2(a)}})\). Key steps include manipulating the terms by factoring out \( v_0^2 \sin(2a)/2g \) and balancing the equation through trigonometric identities.

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Lisa...
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Why does
((v0^2 sin (2a))/2g) + ((v0 cos a)/g) sqrt(v0^2 ((sin^2)a) +hg) equal
((v0^2 sin (2a))/2g) (1+ sqrt(1+((2hg)/(vo^2) (sin^2)a))) ?
 
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Multiply the right-hand term on the lhs by sin(a) and divide the sq root term by sin²(a) to balance out.
Take out the Vo term from inside the sq root and you should have Vo².sin(2a)/2g as a commom factor.

Let me guess: this to find the angle of elevation for maximum range when firing off a cliff of height h, yes ?
 
Last edited:
Exactly! Wow thanks a hell of a lot!
 

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