A skydiver jumps from an airplane at an altitude of 2,500 ft. He falls under the force of gravity until he opens his parachute at an altitude of 1,000 ft. Approximately how long does the jumper fall before he opens his chute?

Answer 1

d=0.5*g*t^2
therefore, t^2= d/ (0.5*g)

where:
d= 2500 ft
g = 32.2 ft /sec^2

solving t:

t^2 = 2500/ (0.5*32.2)

t^2 = 155.28 sec^2

t= 12.46 sec

Answer 2

He only falls 1500 ft before the chute opens. So, change d to 1500 in the above solution and then rework the numbers.

Answer 3

the answer is chicken

Answer 4

A skydiver jumps from an airplane at an altitude of 2,500 ft. He falls under the force of gravity until he opens his parachute at an altitude of 1,000 ft. Approximately how long does the jumper fall before he opens his chute?

Answer 5

Gooday, after banging my head against the wall for 2 hours, not being able to find an explanation, I finally got the answer and I'm content to share my process. The equation utilized to solve this problem is h(t) = at^2 + vt + h(0) (h(t) is the height given the time, a is the acceleration due to gravity, v is initial velocity, and h(0) is the initial height)
The original problem from edgenuity shows an equation with a being 16ft/s 🥴. Acceleration due to gravity is 32ft/s on earth.
Our equation to solve for t looks like: -1500 = -32t^2 + 0v + (2500-1000)
h(t) is -1500 because we had a downwards (negative) displacement of 1500
0v parachuter started off with no velocity and just dropped
-32 <--- negative cuz we falling
Then: -1500 = -32t^2+1500
-3000 = -32t^2
93.75 = t^2
√(93.75) = √(t^2)
t = 9.682458... 😌👍