Answer:
0.293 s
Explanation:
Using equations of motion,
y = 66.1 cm = 0.661 m
v = final velocity at maximum height = 0 m/s
g = - 9.8 m/s²
t = ?
u = initial takeoff velocity from the ground = ?
First of, we calculate the initial velocity
v² = u² + 2gy
0² = u² - 2(9.8)(0.661)
u² = 12.9556
u = 3.60 m/s
Then we can calculate the two time periods at which the basketball player reaches ths height that corresponds with the top 10.5 cm of his jump.
The top 10.5 cm of his journey starts from (66.1 - 10.5) = 55.6 cm = 0.556 m
y = 0.556 m
u = 3.60 m/s
g = - 9.8 m/s²
t = ?
y = ut + (1/2)gt²
0.556 = 3.6t - 4.9t²
4.9t² - 3.6t + 0.556 = 0
Solving the quadratic equation
t = 0.514 s or 0.221 s
So, the two time periods that the basketball player reaches the height that corresponds to the top 10.5 cm of his jump are
0.221 s, on his way to maximum height and
0.514 s, on his way back down (counting t = 0 s from when the basketball player leaves the ground).
Time spent in the upper 10.5 cm of the jump = 0.514 - 0.221 = 0.293 s.