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

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Question 167431This question is from textbook Algebra and Trigonometry
: Luis wanted to throw an apple to Kim, who was on a balcony 40ft above him, so he tossed it upward with an intial speed of 56ft/s. Kim missed it on the way up, but then caught it on the way down. How long was the apple in the air? This question is from textbook Algebra and Trigonometry

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
Ok, start with the general function of the height (h) of an object propelled upwards from an initial height h%5B0%5D with an initial velocity of v%5B0%5D.
10 h%28t%29+=+-16t%5E2%2Bv%5B0%5Dt%2Bh%5B0%5D
In this problem:
h%5B0%5D+=+0 because Luis is on the ground when he throws the apple up to Kim.
v%5B0%5D+=+56ft/sec. This is given in the problem.
You want to find the time (t) at which the apple is at a height of 40 ft.
There will, of course, be two such times, one when the apple is on the up to Kim and the second time (later) when the apple is caught by Kim on the down.
It's the time on the way down that will give the answer to the problem.
So we let h(t) equal to 40 ft. and solve for the two times (t).
40+=+-16t%5E2%2B56t Rewrite into the standard form for quadratic equations.
-16t%5E2%2B56t-40+=+0 Divide through by -8 to simplify the calculations a bit.
2t%5E2-7t%2B5+=+0 Factor the trinomial.
%282t-5%29%28t-1%29+=+0 Applying the zero product rule, we get:
t+=+5%2F2secs. and t+=+1 sec.
So the time on the way up is 1 second but Kim missed this opportunity!
The time on the way down is 2.5 seconds when Kim caught the apple.
So the apple was in the air for 2.5 seconds.