SOLUTION: An Olympic ski jumper skis off a spring board. Her height h, in meters, above the ground, t seconds after she jumps, is given by h(t)= -4.9t^2 + 20t +15. Algebraically determine th

Question 395288: An Olympic ski jumper skis off a spring board. Her height h, in meters, above the ground, t seconds after she jumps, is given by h(t)= -4.9t^2 + 20t +15. Algebraically determine the approximate instantaneous rate of change in her height at 2 seconds and describe what is happening at that time.
Found 2 solutions by ankor@dixie-net.com, richard1234:
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
An Olympic ski jumper skis off a spring board. Her height h, in metres, above
the ground, t seconds after she jumps, is given by h(t)= -4.9t^2 + 20t +15.
Algebraically determine the approximate instantaneous rate of change in her
height at 2 seconds and describe what is happening at that time.
:
A good way to see what is happening is to graph this equation

:
It looks like it is at max height at 2 sec,
prove that with the axis of symmetry formula x = -b/(2a), where a = -4.9, b=20
x =
x =
x = 2.04 sec it is at max height
:
At this point it is changing from an upward path to a downward path
There is no change in height at 2 sec
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Note that t = 2 is fairly close to the vertex which occurs at seconds. Therefore the velocity is close to 0 m/s.

To determine an exact value, if we have , then , and the velocity at time t = 2 is m/s, so the rate of change in height is .4 meters per second (upward since positive numbers are chosen to represent upward for this particular problem).
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