Askydiver jumps out of an airplane. after 0.8 seconds, she has fallen 100 feet. after 3.1 seconds, she has fallen 500 feet. emtiaz says that the skydiver should fall about 187.5 feet in 1.5 seconds. is his answer reasonable? explain.

U = the initial vertical velocity of the skydiver.
g = 32 ft/s², acceleration due to gravity.

Ignore air resistance.
The distance fallen after t seconds is
s = (u ft/s)*(t s) + (1/2)*(32 ft/s²)*(t s)² = ut + 16t²

When t = 0.8 s,
s = 0.8u + 16*0.8² = 0.8u + 10.24
Because s = 100 ft, therefore
u = (100 - 10.24)/0.8 = 112.2 ft/s

When t = 3.1 s,
s = 112.2*3.1 + 16*3.1² = 501.6 ft
This compares favorably with 500 ft.

When t = 1.5 s, then
s = 112.2*1.5 + 16*1.5² = 204.3 ft
This disagrees with the specified value of 187.5 ft

The answer is not reasonable.
Reason:
The answer does not obey the equation
s = u + (1/2)gt²