SOLUTION: suppose that an object is thrown into the air with an intial upward velocity of vo meters per second from a height ho meters above the ground. Then t seconds later, its height h(t)

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: suppose that an object is thrown into the air with an intial upward velocity of vo meters per second from a height ho meters above the ground. Then t seconds later, its height h(t)      Log On


   

Question 171930: suppose that an object is thrown into the air with an intial upward velocity of vo meters per second from a height ho meters above the ground. Then t seconds later, its height h(t) meters above the ground is modeled by the function
h(t)= -4.9 t^2 + vot + ho. (this model dosent' account for resistance)
given that information, the question is:
One half second after springing from a high diving board, a diver reaches her highest point above the water, 4.225 m. If the diving board is 3 meters above the water, how long is the diver in the air?
I would really appreciate any help I can recieve on this problem.
please help soon, thanks!
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
suppose that an object is thrown into the air with an intial upward velocity of
vo meters per second from a height ho meters above the ground.
Then t seconds later, its height h(t) meters above the ground is modeled by the function
h(t)= -4.9 t^2 + vot + ho.
:
given that information, the question is:
One half second after springing from a high diving board, a diver reaches her
highest point above the water, 4.225 m.
If the diving board is 3 meters above the water, how long is the diver in the air?
:
In the above equation we know: t=.5; ho=3; h(t) = 4.225
Complete the equation by finding vo
-4.9(.5^2) + .5vo + 3 = 4.225
-4.9(.25) + .5vo = 4.225 - 3
-1.225 + .5vo = 1.225
.5vo = 1.225 + 1.225
.5vo = 2.45
vo = 2(2.45)
vo = 4.9 m/sec
:
When she strikes the water h(t) = 0, find the value of t when this happens
-4.9t^2 + 4.9t + 3 = 0
:
Solve this equation using the quadratic formula
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
In this equation x=t; a=-4.9; b=4.9; c=3
t+=+%28-4.9+%2B-+sqrt%284.9%5E2+-+4%2A-4.9%2A3+%29%29%2F%282%2A-4.9%29+
:
t+=+%28-4.9+%2B-+sqrt%2824.01+%2B+58.8+%29%29%2F%28-9.8%29+
:
t+=+%28-4.9+%2B-+sqrt%2882.81%29%29%2F%28-9.8%29+
We want the positive solution here
t+=+%28-4.9+-+9.1%29%2F%28-9.8%29
t = +1.43 seconds the diver is in the air
:
The equation looks like this graphically
+graph%28+300%2C+200%2C+-.5%2C+2%2C+-3%2C+5%2C+-4.9x%5E2%2B4.9x%2B3%29+
Note the highest point is 4.2 m at .5 seconds and crosses the x axis about +1.4