SOLUTION: A Little League baseball player throws a ball upward. The height h of the ball (in feet) t seconds after the ball is released is given by the quadratic equation
h = -16t2 +30t +4
Algebra ->
Customizable Word Problem Solvers
-> Misc
-> SOLUTION: A Little League baseball player throws a ball upward. The height h of the ball (in feet) t seconds after the ball is released is given by the quadratic equation
h = -16t2 +30t +4
Log On
Question 652685: A Little League baseball player throws a ball upward. The height h of the ball (in feet) t seconds after the ball is released is given by the quadratic equation
h = -16t2 +30t +4
a) How long does it take the ball to reach a height of 18ft?
b) How long does it take the ball to hit the ground?
In the future, please do not waste tutors' time by submitting the same problem twice. Here is the answer I gave you before.
a. Substitute the height value 18 for the variable representing height, , then put the equation into standard form and solve the factorable quadratic. You will have two values, one is the time it will reach 18 feet on the way up and the other is the time it will be at 18 feet a second time on the way down.
b. Substitute the height value 0 for the variable representing height (the height of the ground is zero), , then solve the factorable quadratic. Discard the negative root because you don't care what happened before he threw the ball. (Do you?!)
John
My calculator said it, I believe it, that settles it
You can put this solution on YOUR website! A Little League baseball player throws a ball upward. The height h of the ball (in feet) t seconds after the ball is released is given by the quadratic equation
h = -16t2 +30t +4
a) How long does it take the ball to reach a height of 18ft?
set h to 18 and solve for t:
h = -16t^2 +30t +4
18 = -16t^2 +30t +4
0 = -16t^2 +30t - 14
0 = 16t^2 -30t + 14
0 = 8t^2 -15t + 7
0 = 8t^2 -8t -7t + 7
0 = (8t^2 -8t) - (7t - 7)
0 = 8t(t-1) - 7(t-1)
0 = (t-1)(8t-7)
t = {1, 7/8}
t = {1, 0.875}
it reaches 18 ft twice. Once at 0.875 sec (going up) and again at 1 sec (going down).
.
b) How long does it take the ball to hit the ground?
set h to 0 and solve for t:
h = -16t^2 +30t +4
0 = -16t^2 +30t +4
0 = 16t^2 -30t -4
0 = 8t^2 -15t -2
0 = 8t^2 -16t+t -2
0 = (8t^2-16t) + (t-2)
0 = 8t(t-2) + (t-2)
0 = (t-2)(8t+1)
t = {-1/8, 2}
we can throw out the negative solution leaving:
t = 2 seconds
