# algebra 2

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?

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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

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

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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?

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