Surprisingly, very few athletes can jump more than 2.6 ft (0.8 m) straight up. Use d = 1/2 gt2 and solve for the time one spends moving upward in a 2.6 foot vertical jump. Then double it for the "hang-time" -- the time one's feet are off the ground.
12.5
To solve for the time spent moving upward in a 2.6 foot vertical jump using the equation d = 1/2 gt^2, we need to rearrange the equation to solve for time (t).
The equation for the distance (d) can be written as:
d = 1/2 gt^2
We want to find the time (t), so we rearrange the equation:
2d/g = t^2
Now, substitute the given values into the equation:
2 * 2.6 ft / (32.2 ft/s^2) = t^2
Simplify the equation:
5.2 ft / 32.2 ft/s^2 = t^2
Divide to solve for t^2:
t^2 = 0.1615 s^2
Now take the square root of both sides to find t:
t ≈ √0.1615 s ≈ 0.402 s
This gives us the time spent moving upward in a 2.6 foot vertical jump. To calculate the hang-time (time one's feet are off the ground), we simply double this value.
Hang-time = 2 * 0.402 s ≈ 0.804 s
Therefore, in a 2.6 foot vertical jump, the time spent moving upward would be approximately 0.402 seconds, and the hang-time would be approximately 0.804 seconds.