Answer :
The minimum initial velocity u = 58.3°
(v0) min = 9.76 m/s
Explanation:
Coordinate System: The coordinate system will be set so that its origin coincides with the ball’s initial position.
x-Motion: Here (v0) = v0 cos u x0 and x = 6 m Thus,
(+) x = [tex]x_{0}[/tex] + (v0)xt
6 = 0 + (v0 cos u)t
t = 6/v0 cos u (1)
y-Motion: Here (v0)x = v0 sin u, ay = -g = -9.81m/s and y0 = 0
( A + c B) y = y0 + (v0)y t +1/2 [tex]a_{y}[/tex][tex]t^{2}[/tex]
3 = 0 + v0 (sin u)t + 1/2 (-9.81) [tex]t^{2}[/tex]
3 = v0 (sin u)t - 4.905[tex]t^{2}[/tex] (2)
Substituting Eq. (1) into Eq.(2) yields
v0 = [tex]\sqrt[2]{\frac{58.86}{sin 2u - cos2 u} }[/tex] (3)
From Eq. (3), we notice that is minimum when is
maximum. This requires [tex]\frac{df(u)}{du}[/tex]
[tex]\frac{df(u)}{du}[/tex] = 2 cos2u + sin2u = 0
tan2u = -2
2u = 116.57°
u = 58.28° = 58.3°
Substituting the result into Eq.(2), we have
(v0)min = [tex]\sqrt[2]{\frac{58.86}{sin 116.57° - cos2 58.28°} }[/tex]
(v0)min = 9.76 m/s
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