Question

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? For this quadratic model we will let the y-axis be the axis of symmetry.

Answer 1

Answer:

9.7 seconds to the nearest tenth.

Step-by-step explanation:

He falls for a distance of 1,500 ft.

We use the equation of motion  s = 1/2 * 32 * t^2   where s = distance and t = time.

1500 = 16 * t^2

t^2  = 93.75

t =  9.7 seconds answer.

Answer 2

Answer:

9.7

Step-by-step explanation:

minus 1,000 ft (the height he opened his parachute) from 2,500 (the height of which he started)

This will give you the distance of his fall

1,500 ft.

Then, use the equation of motion which is s = 1/2 * 32 * t^2  

where s represents distance and t is time

1500 = 16 * t^2

t^2  = 93.75

Then simply round to the nearest tenth and it will be on the test :)

I really just ignored the equation it gives to solve it sorry.