SOLUTION: Hello! I'm having trouble with a Quadratic Equation word problem, quite frankly, I don't even know where to start. I know the formula, so I'm confident that I can solve it once it'
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Question 763472: Hello! I'm having trouble with a Quadratic Equation word problem, quite frankly, I don't even know where to start. I know the formula, so I'm confident that I can solve it once it's in an equation form, but I just can't seem to get there. The question is:
Jasmine is a competitive diver. In a dive off an 8m platform, she reaches a maximum height of 8.4m after 0.30 seconds. How long does it take her to reach the water?
Thank's so much for the help, I really appreciate it!
Found 2 solutions by ankor@dixie-net.com, rothauserc:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
Jasmine is a competitive diver.
In a dive off an 8m platform, she reaches a maximum height of 8.4m
after 0.30 seconds.
How long does it take her to reach the water?
:
The question is how long does it take to fall 8.4 meters
The water is 0, therefore:
-4.9t^2 + 8.4 = 0
-4.9t^2 = -8.4
t^2 =
t^2 = +1.71
t =
t = 1.3 seconds which is 1.6 seconds after she leaves the board
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
The quadratic equation to use is
s(t) = -4.9*t^2 +v(0)*t +h(o)
note that we use 4.9 since we are working in meters, also h(o) = 8 meters
==============================================================================
now we need to calculate v(o) which is the launch velocity and
we use rate * time = distance to calculate the rate which is v(0), so
v(0) * .30 seconds = .4 meters and
v(0) = .4 / .30 = 1.33 meters/sec
======================================================================
now returning to the quadratic above we have
0 = -4.9*t^2 +1.33*t + 8
solve for t using the quadratic formula
t = (-b+sqrt(b^2-4ac)) / 2a = (-1.33 +sqrt(1.33^2 -4*(-4.9)*8)) / 2*-4.9 = -1.15
t = (-b-sqrt(b^2-4ac)) / 2a = (-1.33 +sqrt(1.33^2 -4*(-4.9)*8)) / 2*-4.9 = 1.42
therefore she reaches the water in 1.42 seconds
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