Answer to Question #1853 in Mechanics | Relativity for BALACHANDRA K
H = h0 + voyt - gt2/2
S = vox t
Where vox = Vo cos(α), voy = Vo sin(α), ho = 0, H = 0.
0 = Vo sin(α)t - gt2/2; t =0; t = 2Vo sin(α) / g
S = Vo cos(α)t
S = 2Vo cos(α)Vo sin(α) / g = Vo sin (2α) / g , the maximum value of sin(2α) is 1 at 2α = 90, hence
α = 45 degrees.
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