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

Question

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

Answer 1

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)

Answer 2

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.

Answer 3

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)

Answer 4

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

Answer 5

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.

Answer 6

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

Answer 7

6.6 seconds and 6.9 seconds, respectively.