This question involves the concepts of the law of conservation of energy, Pythagora's Theorem, kinetic energy, and potential energy.
The rope makes an angle of "26.2°" with the vertical when the ball's kinetic energy is halved.
First, we will find the initial kinetic energy of the ball:
[tex]K.E = \frac{1}{2}mv^2\\\\K.E=\frac{1}{2}(m)(1.1\ m/s)^2\\\\K.E = (0.605\ m^2/s^2) m[/tex]
Now, half of this kinetic energy is left at the final point. Therefore, according to the law of conservation of energy, the other half must be converted to the potential energy of the ball:
[tex]P.E=\frac{1}{2}K.E\\mgh'=\frac{1}{2}(0.605\ m^2/s^2)m\\\\h'=\frac{0.3025\ m^2/s^2}{9.81\ m/s^2}\\\\h' = 0.031\ m[/tex]
This h' is the height of the ball above the ground the final position, as shown in the picture attached.
Applying Pythagora's Theorem to the right angled triangle shown in the attached picture we have:
[tex]Cos\theta = \frac{Base}{Hypotenuse}=\frac{h}{l}[/tex]
where,
θ = angle of rope with vertical = ?
l = legth of rope = 0.3 m
h = l - h' = 0.3 m - 0.031 m = 0.27 m
Therefore,
[tex]Cos\theta=\frac{0.27\ m}{0.3\ m}\\\\\theta = Cos^{-1}(0.89721)[/tex]
θ = 26.2°
Learn more about the law of conservation of energy here:
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