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) 1.01 s

(B)

Explanation:

horizontal distance (L) = 3 m

vertical height (h) = 1.25 m

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

(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.

         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

      1.25 = 0.5 x 9.8 x

      = 1.25/(0.5x9.8)

      t =

      t = 0.505 s

       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

answer is c

explanation: