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Calculus/projectiles

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Expert: - 12/6/2008


Question
Dear Sir : I have been asked to find the angle that gives the maximum length of arc in ( horizontal) projectiles  . can you help please show steps .
Thank you

Answer
Take v as the velocity and we have the distance is cos(α)vt.
The speed at which the projectile is going upwards speed is sin(α)v.

Time in the air is found when sin(α)vt – atē/2 = 0 (from physics)

Solving this leads to t(sin(α)v – at/2) = 0, so we can say that the time in the air t1=-2sin(α)v/a (since when t1=0 we have just started).

Note that the letter α is the only variable in this equation.\

The horizontal distance travelled is –2sin(α)cos(α)vē/a.

Take the derivative of this equation with respect to α and you get
(-2vē/a)(sinē(α)- cosē(α))=0.

Note that the term out front is a constant that is never 0.

To solve sinē(α)=cosē(α), since it is known that α is between 0° and 90°, we must have α = 45°, for that's the only angle where
sin(α) = cos(α).

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