An acrobat is launched from a cannon at an angle of 60 degrees above the horizontal. The acrobat is caught by a safety net mounted horizontally at the height from which he was initially launched, suppose the acrobat is launched at a speed of 26 m/s, how long does it take before he reaches his maximum height? How long does it take in total for him to reach a point halfway back down to the ground

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]