Question 992241: A person standing close to the edge on the top of a 270-foot building throws a baseball vertically upward. The quadratic equation
\qquad\qquad h = -16 t^2 + 160 t + 270
models the ball's height \, h \, above the ground in feet, t seconds after it was thrown.
How high is the ball after 5 seconds?
??feet.
How many seconds does it take until the ball finally hits the ground? Round to the nearest tenth of a second.
?? seconds.
(I plugged in 5, then set it equal to 0, none of it worked, I'm probably doing it wrong)
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! h = -16 t^2 + 160 t + 270
to find the height of the ball after 5 seconds substitute 5 for t
h = -16(5^2) + (160*5) + 270
h = 670 feet high after 5 seconds
*****************************************************************
to find when the ball hits the ground, set h = 0 and solve the quadratic for t
-16 t^2 + 160 t + 270 = 0
multiply both sides of = by -1
16t^2 - 160t - 270 = 0
use quadratic formula to solve for t
t = (160 + sqrt(160^2 - (4*16*(-270)))) / (2*16) = 11.471089553 approx 11.5
t = (160 - sqrt(160^2 - (4*16*(-270)))) / (2*16) = −1.471089553 approx -1.5
****************************************************************************
we want the positive value for t
t = 11.5 seconds for the ball to hit the ground
|
|
|
