SOLUTION: Malik popped a ball staight up with an initial upward velocity of 45 feet per second. the height h, in feet, of the ball above the ground is modeled by the equation h = 2 + 45t - 1

Algebra ->  Polynomials-and-rational-expressions -> SOLUTION: Malik popped a ball staight up with an initial upward velocity of 45 feet per second. the height h, in feet, of the ball above the ground is modeled by the equation h = 2 + 45t - 1      Log On


   

Question 608446: Malik popped a ball staight up with an initial upward velocity of 45 feet per second. the height h, in feet, of the ball above the ground is modeled by the equation h = 2 + 45t - 16t^2. How long was the ball in the air if the catcher catches the ball when it is 2 feet above the ground?
Found 2 solutions by nerdybill, lwsshak3:
Answer by nerdybill(7384) About Me  (Show Source):
You can put this solution on YOUR website!
Given:
h = 2 + 45t - 16t^2
set h to 2 and solve for t:
2 = 2 + 45t - 16t^2
0 = 45t - 16t^2
0 = t(45 - 16t)
.
t = {0,45/16}
t = {0,2.81}
answer: 2.8 seconds

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
Malik popped a ball staight up with an initial upward velocity of 45 feet per second. the height h, in feet, of the ball above the ground is modeled by the equation h = 2 + 45t - 16t^2. How long was the ball in the air if the catcher catches the ball when it is 2 feet above the ground?
**
t=seconds ball is in the air after it is thrown
h=height ball is in the air after t seconds
h = 2 + 45t - 16t^2
given height is 2 ft
2 = 2 + 45t - 16t^2
0=45t-16t^2
16t^2-45t=0
t(16t-45)=0
t=0 (ball is 2 ft above the ground before it is thrown)
or
16t-45=0
t=45/16≈2.81 sec (ball is 2 ft above the ground on its way down)