SOLUTION: Jack wanted to throw an apple to Lauren, who was on a balcony 40 feet above him, so he tossed it upward with an initial speed of 56 ft/s. Lauren missed it on the way up, but then

Question 372797: Jack wanted to throw an apple to Lauren, who was on a balcony 40 feet above him, so he tossed it upward with an initial speed of 56 ft/s. Lauren missed it on the way up, but then caught it on the way down. How long was the apple in the air?
Would I use the equation h(t) = -16t^2 + 56t +40? (where h = height and t = time) I'm not very good at solving quadratic word problems :(
Answer by nerdybill(7384) (Show Source): You can put this solution on YOUR website!
Well, you're close
The general equation is
h(t) = -(1/2)gt^2 + Vot + h(0)
where
g is gravity 32 ft/sec^2
Vo is initial velocity
h(0) is initial height of Jack (not Lauren)
.
So, plugging in what we know
h(t) = -(1/2)(32)t^2 + 56t + 0
which reduces to
h(t) = -16t^2 + 56t
.
So, since Lauren is 40 feet up -- set h(t) to 40 and solve for t:
h(t) = -16t^2 + 56t
40 = -16t^2 + 56t
0 = -16t^2 + 56t - 40
multiply both sides by -1:
0 = 16t^2 - 56t + 40
divide both sides by 8:
0 = 2t^2 - 7t + 5
0 = (2t-5)(t-1)
So,
t = {1, 5/2}
1 sec represents the time the apple reaches 40 ft (going up)
So
5/2 secs or
2 and 1/2 secs (is your solution)

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