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.

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sofiasilva sofiasilva · Answer 1 · 2020-06-16T13:45:55+02:00

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.

priyankapandeyVT priyankapandeyVT · Answer 2 · 2022-07-23T15:17:44+02:00

The required time taken by the skydiver from the jump to before it opens chute is 9.7 second.

A skydiver jumps at 2500 ft. and opens it chute at 1000 ft.
Time interval for chute to be not opened to be determined.

Gravity is gravitational force exerted by the earth.

Here,
Distance travel when chute is not opened = 2500 - 1000.
= 1500 ft.

Since initial velocity would be zero for free falling bodies. so 2nd equation of motion.

s = 1/2gt^2
where s = displacement and g = 32 feet/sec.
1/2 x 32 x t^2 = 1500
t^2 = 1500/16
t = 9.7 sec.

Thus, the required time taken by the skydiver from the jump to before it opens chute is 9.7 second.

Learn more about gravity here:
https://brainly.com/question/4014727

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