Answer:
The acrobat takes approximately 2.30 seconds to reach the maximum height. The total time for the acrobat to reach a point halfway back down to the ground is approximately 3.92 seconds.
Explanation:
1. Calculate the initial vertical velocity [tex](\(v_{y0}\))[/tex]:
[tex]\[ v_{y0} = v_0 \sin(\theta) \][/tex]
where [tex]\(v_0 = 26 \, \text{m/s}\)[/tex] (the initial launch speed) and [tex]\(\theta = 60^\circ\)[/tex] (the launch angle). Convert [tex]\(\theta\)[/tex] to radians for the calculation.
2. Determine the time to reach maximum height [tex](\(t_{\text{max height}}\))[/tex]:
[tex]\[ t_{\text{max height}} = \frac{v_{y0}}{g} \][/tex]
where \(g = 9.81 \, \text{m/s}^2\) (the acceleration due to gravity).
3. Calculate the maximum height [tex](\(h_{\text{max}}\))[/tex]:
[tex]\[ h_{\text{max}} = \frac{v_{y0}^2}{2g} \][/tex]
4. Calculate the time to fall from maximum height to half of maximum height [tex](\(t_{\text{half fall}}\)[/tex]):
[tex]\[ t_{\text{half fall}} = \sqrt{\frac{h_{\text{max}}}{g}} \][/tex]
5. Find the total time from launch to the point halfway back down to the ground [tex](\(t_{\text{total halfway}}\)[/tex]):
[tex]\[ t_{\text{total halfway}} = t_{\text{max height}} + t_{\text{half fall}} \][/tex]
