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OK, so a person jumps off from a 10-meter springboard...h in meters and t in seconds, after the diver leaves the board her height above the water is given by
-4.9t^2+8t+10

F(x)=-4.9(t+.8)^2+13.1....vertex form

BUT

How long does it take the diver to reach the water

At what time is the diver EXACTLY 11m above the water

For how long is the diver 11m above the water

PLEASE try to use a graphing calculator and show the strokes because I don't know how to do these...and include ur answers so I can double check...

THANKS!!!

2007-11-28 20:30:15 · 1 answers · asked by A A 1 in Science & Mathematics Mathematics

1 answers

h = - 4.9t^2 + 8t + 10
h = - 4.9(t^2 - (8/4.9)t) + 10
h = - 4.9(t^2 - (8/4.9)t + (4/4.9)^2) + 16/4.9 + 10
h = - 4.9(t - 4/4.9)^2 + 65/4.9 in vertex form

How long does it take the diver to reach the water?
4.9(t - 4/4.9)^2 = 65/4.9
(t - 4/4.9)^2 = 65/4.9^2
t - 4/4.9 = (√65)/4.9
t = (4 + √65)/4.9
t ≈ 2.462 s

At what time is the diver EXACTLY 11m above the water?
11 = - 4.9(t - 4/4.9)^2 + 65/4.9
4.9(t - 4/4.9)^2 = 11.1/4.9
t - 4/4.9 = ± (√11.1)/4.9
t = (4 - (√11.1)/4.9, (4 + (√11.1)/4.9
t ≈ 0.136 s, 1.496 s

For how long is the diver 11m or more above the water?
1.360 s

Sorry, but I don't own a graphing calculator.

2007-11-28 21:36:09 · answer #1 · answered by Helmut 7 · 0 0

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