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# A smoke jumper jumps from a plane at 1700 feet above ground. The function y=-16t^2+1700 gives the jumper's he

A smoke jumper jumps from a plane at 1700 feet above ground. The function y=-16t^2+1700 gives the jumper's height y in feet at t seconds.

How long is the jumper in free fall if the parachute opens at 1000 feet? What about at 940 feet???

please help

### 7 Answers

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just substitute the height the shute opens at for y and then solve for t to get the time.

1000 = -16t^2 + 1700

t = sqrt(43.75)

940 = -16t^2 + 1700

t = sqrt(47.5)

• y = -16t²+1700

For height 1000 metes:

1000 = -16t²+1700

1000 + 16t² = +1700

16t² = +1700 - 1000

16t² = 700

t² = 700 /16

t² = √[700 /16]

t = 6∙614 378......

t ≈ 6∙61 sec.

y = -16t²+1700

For height 940 metes:

940 = -16t²+1700

940 + 16t² = +1700

16t² = +1700 - 940

16t² = 760

t² = 760 /16

t² = √[700 /16]

t = 47∙5

t ≈ 6∙892 02....

t ≈ 6∙89 sec.

• The function for the time in freefall is: y=-16t^2+1700

In the two situations, the height (y) will be equal to 940, and a 1000.

Plug in the values for y:

1000=-16t^2+1700

-700=-16t^2

-175/4=-t^2

t^2=175/4

t= sqr (175)/2

t=5 sqr 7/2, or 2.5 sqr. 7

The next value: 940

940=-16t^2+1700

-760=-16t^2

t^2=760/16

t=2 sqr.(190)/4

t=0.5 sqr. (190)

• The equation for free fall

y=-16t^2+1700

1] y=1000

1000=-16t^2+1700

t^2=700/16

T=6.6143sec

2]y=940

similarly

t^2=760/16

t=6.89sec

• If the parachute opens at 1000 feet, he is in free-fall for 6.61 seconds. If it opens at 940 feet, he is in free-fall for 6.89 seconds.

it's very simple algebra, just plug in 1000 or 940 for y, then solve the given equation for t.

• y=-16t^2+1700

y1=1700, t1=0

y2=1000, t2=?

1000=-16(t2)^2+1700

(t2)^2 = (1000-1700)/(-16)

t2 = SQRT(43.75) = 6.61 seconds

Repeat with y3=940

• Anonymous
1 decade ago

6.6 seconds and 6.9 seconds, respectively.

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