Answer:
t_up / t_down = 6.83
Explanation:
Find:
Calculate the ratio of the time he is above y_max/2 to the time it takes him to go from the floor to that height.
Solution:
- Compute the velocity v_o at y_max:
v_i^2 = v_f^2 - 2*g*y_max
0 = v_o^2 - 2*g*y_max
v_o = sqrt (2*g*y_max)
- The total time spend by athlete above height y_max / 2 is:
y - y_o = v_o*t_up - 0.5*g*t^2_up
v_o = 0.5*g*t_up
- Equate two equations:
sqrt (2*g*y_max) = 0.5*g*t_up
t_up = 2*sqrt(2*g*y_max) / g
- The total time taken by athlete to reach height y_max / 2 from ground is:
y - y_o = v_o*t_down - 0.5*g*t^2_down
g*t_^2down - 2*v_o*t_down + y_max = 0
- Solve the quadratic and evaluate t_down:
t_down = (v_o +/- sqrt (v^2_o - g*y_max)) / g
Substitute for v_o = sqrt (2*g*y_max)
t_down = (sqrt(2g*y_max) +/- sqrt(g*y_max)) / g
- We will use the minus quantity, because we need the first part of the journey from ground from the two times he passes the height of y_max/2.
Hence,
t_down = (sqrt(2g*y_max) - sqrt(g*y_max)) / g
t_down = (sqrt(g*y_max) / g) * (sqrt(2) - 1)
- Compute the ratio t_up to t_down:
t_up / t_down = 2*sqrt(2*g*y_max) / g * g / (sqrt(g*y_max)*(sqrt(2) - 1)
= 2*sqrt(2) / (sqrt(2) - 1)
= 6.83
