Respuesta :
Answer:
2.4
Explanation:
To find the time above and below ymax/2 we use the following kinematic formula
[tex]v=v_{0}-gt[/tex]
Where
[tex]v[/tex] = Final Velocity
[tex]v_{0}[/tex] = Initial Velocity
[tex]g[/tex] = Gravity
[tex]t[/tex] = Time
Notice that we need to first find the values of velocity to solve this equation. To do that we use the following kinematic formula
[tex]v_{1}^{2} = v_{0}^{2} - 2gh[/tex]
Where
[tex]v_{1}[/tex] = Final Velocity
[tex]v_{0}[/tex] = Initial Velocity
[tex]g[/tex] = Gravity
[tex]h[/tex] = Height
Taking h = ymax/2 we get the following formula
[tex]v_{1}^{2} = v_{0}^{2} - 2g(\frac{y_{max} }{2})[/tex]
To find the value of value of ymax in terms of v0 we use the law of conversation of energy
[tex]KE = PE\\ \frac{1}{2}mv_{0} ^{2} = mgy_{max} \\ y_{max} = \frac{v_{0}^2 }{2g}[/tex]
Now we can substitute the value of ymax
[tex]v_{1}^{2} = v_{0}^{2} - 2g(\frac{v_{0}^2 }{4g})\\ v_{1}^{2} = \frac{v_{0}^2 }{2}\\ v_{1} = \frac{v_{0} }{\sqrt{2} }[/tex]
Now using the original equation to find time, we input the value of V1 and find time in terms of V0
Assuming final velocity v is 0 (since at the top velocity is zero)
[tex]v=v_{0}-gt\\ t = \frac{v_{0}}{g}[/tex]
This gives us total time from bottom to top
To find time from ymax/2 to top we substitute the value of V1
[tex]t_{1} = \frac{v_{1}}{g}\\ t_{1} = \frac{v_{0} }{g\sqrt{2} }[/tex]
The time he is above ymax/2 is nothing more than the difference between t and t1
[tex]t - t_{1} = \frac{v_{0} }{g} (1 - \frac{1}{\sqrt{2}})[/tex]
To find the required ratio we divide t1 by (t - t1)
[tex]\frac{t_{1}}{t - t_{1}} = 2.4[/tex]