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.
: 
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 = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
In this equation x=t; a=-4.9; b=4.9; c=3
{{{t = (-4.9 +- sqrt(4.9^2 - 4*-4.9*3 ))/(2*-4.9) }}}
:
{{{t = (-4.9 +- sqrt(24.01 + 58.8 ))/(-9.8) }}}
:
{{{t = (-4.9 +- sqrt(82.81))/(-9.8) }}}
We want the positive solution here
{{{t = (-4.9 - 9.1)/(-9.8)}}}
t = +1.43 seconds the diver is in the air
:
The equation looks like this graphically
{{{ graph( 300, 200, -.5, 2, -3, 5, -4.9x^2+4.9x+3) }}}
Note the highest point is 4.2 m at .5 seconds and crosses the x axis about +1.4