urprisingly, very few athletes can jump more than 2.2 ft (0.66 m) straight up. Use d = 1/2 gt2 and solve for the time one spends moving upward in a 2.2 foot vertical jump. Then double it for the "hang-time" -- the time one's feet are off the ground.
ho=0
hf=.66m
hf=ho+1/2 9.8 t^2 solve for t, then double it.
Can you please explain how to find time by using step by step.
To calculate the time one spends moving upward in a vertical jump, we can use the equation:
d = (1/2)gt^2
Where:
d is the vertical distance traveled (2.2 ft or 0.66 m),
g is the acceleration due to gravity (approximately 9.8 m/s^2), and
t is the time spent moving upward.
First, let's convert the distance from 2.2 feet to meters:
0.66 m = 2.2 ft × 0.3048 m/ft
0.66 m = 0.67056 m
Now, we can rearrange the equation to solve for t:
t^2 = (2d) / g
t^2 = (2 * 0.67056 m) / 9.8 m/s^2
t^2 = 0.1366464 m / 9.8 m/s^2
t^2 = 0.0139647 s^2
To solve for t, we take the square root:
t = √(0.0139647 s^2)
t ≈ 0.1182 s
The time spent moving upward in a 2.2-foot jump is approximately 0.1182 seconds.
To find the "hang-time," which is the time one's feet are off the ground, we double the time spent moving upward:
Hang-time = 2 * t
Hang-time ≈ 2 * 0.1182 s
Hang-time ≈ 0.2364 s
Therefore, the "hang-time" in a 2.2-foot vertical jump is approximately 0.2364 seconds.