Calculate the hang time of a person who moves 3m horizontally during a 1.25-m high jump. What is the hang time when the person moves 6m horizontally during this jump?

(A) hang time refers to how long the person remains in the air, and is the summation of the time it took attain maximum height and the time it takes to get back to the ground.

it is going to take the same amount of time to attain maximum height and to return to the ground.
from s = ut + 0.5g[tex]t^{2}[/tex] we can get the time taken to attain maximum height.

where s = vertical distance = 1.25 m

u = initial velocity = 0 (since body starts from rest)

g = acceleration due to gravity = 9.8 m/s^{2}

t == time taken to get to maximum height

the equation now becomes s = 0.5g[tex]t^{2}[/tex]

1.25 = 0.5 x 9.8 x [tex]t^{2}[/tex]

[tex]t^{2}[/tex] = 1.25/(0.5x9.8)

t = [tex]\sqrt{0.255}[/tex]

t = 0.505 s

now that we have the time taken to attain maximum height which is also equal to the time it takes to get back to the ground.

total time = time taken to attain maximum height + time it takes to get back to the ground.

total time = 0.505 + 0.505 = 1.01 s

(B) when horizontal distance = 6 m , the hang time would remain the same, because the horizontal component does not affect the vertical component . Also we can see that from the equation used in part A above the horizontal component was not used to solve for the time. Therefore hang time = 1.01 s