My question is straight out of the textbook, and I'm struggling to understand it.
Surprisingly, very few athletes can jump more than 2 feet
(0.6 m) straight up. Use d = 1/2gt2 to solve for the time
one spends moving upward in a 0.6-m vertical jump.
Then double it for the “hang time”—the time one’s feet
are off the ground.
If g = 9.81 m/s^2
then
.6 = (1/2)(9.81) t^2
t^2 = 0.1223
t = 0.35 seconds up
2 t = 0.70 seconds in the air
reason
F = m a
a = -g = -9.81
then
v = Vi - 9.81 t where v is speed up and Vi is initial speed leaving ground
then
H = Vi t - (9.81/2) t^2 where H is the amount up in tie t
at top v = 0
so
Vi = 9.81 t at the top
thenat top
H = 9.81 t^2 - (9.81/2) t^2
= (1/2) (9.81) t^2 like we said
To solve the problem, you need to use the formula d = 0.5 * g * t^2, where d is the vertical distance, g is the acceleration due to gravity, and t is the time.
In this case, the vertical distance is 0.6 m, so you can write the equation as:
0.6 = 0.5 * g * t^2
To solve for the time spent moving upward in a 0.6 m vertical jump, you can rearrange the equation:
t^2 = (2 * 0.6) / g
t^2 = 1.2 / g
t = sqrt(1.2 / g)
Next, you need to double the time to calculate the "hang time" when the feet are off the ground. So the final equation for hang time would be:
hang time = 2 * sqrt(1.2 / g)
To find the actual numerical value of the hang time, you need to know the value of the acceleration due to gravity. On Earth, the value of g is approximately 9.8 m/s^2. So you can substitute this value into the equation:
hang time = 2 * sqrt(1.2 / 9.8)
Simplifying this expression will give you the time spent with feet off the ground for a 0.6 m vertical jump.