Answer to Question #2904 in Mechanics | Relativity for pradeep
Let's denote the distance as S and write the equations of the motion under the gravitational acceleration:
H = h0 + Vy*t - gt2/2
S = Vx*t.
Here Vx = V0*cos(α), Vy = V0*sin(α), h0 = 0, H = 0.
0& =& Vo sin(α)t - gt²/2; &
t =0;& t = 2V0*sin(α)/g
S = V0*cos(α)t &
S& = 2V0*cos(α)V0*sin(α)/g = V0*sin(2α)/ g , the maximum value of& sin(2α) is 1 at 2α = 90, hence α = 45 degrees.
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