Question #ebb84

1 Answer
Eddie
Apr 25, 2017

At max height the projectile is travelling horizontally.

Assuming no resistance, and for launch veclocity vec v_o→vo at angle thetaθ to horizontal, it's horizontal velocity throughout is abs (vec v_o) cos theta \ vece_x.

So we can say that:

abs (vec v_o) = 5 abs (vec v_o) cos theta

implies theta = cos^(-1) \ 1/5

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