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Help with Algebra 2 problem, please...not sure if I did it right! Thanks!?
While Don is grounded his friend Joe brings him a video game. Don stands at his bedroom window, & Joe stands directly below the window. If Joe throws the game to Don with an initial velocity of 35 feet per second, an equation for the height "h" feet of the game after "t" seconds is:
h = -16t^2 + 35t + 5.
a. If the window is 25 feet above the ground, will Don have 0,1, or 2 chances to catch the game?
b. If Don is unable to catch the game, when will it hit the ground?
1 Answer
- Anonymous1 decade agoFavorite Answer
(i) In order to see how many chances Don has to catch the game, we need to see how many times the game is 25 feet above the ground. This occurs when h = 25.
So we have:
25 = -16t^2 + 35t + 5
==> 16t^2 - 35t + 20 = 0.
Using the discriminant:
b^2 - 4ac = (-35)^2 - 4(16)(20) = -55 < 0.
Since this is negative, the game never gets to him. Thus, he has no chance at getting the game.
(ii) The game will hit the ground when h = 0. This gives:
0 = -16t^2 + 35t + 5
==> 16t^2 - 35t - 5 = 0.
There are two solutions to do this: one negative and one positive. Use the Quadratic Formula to obtain the two values of t and reject the negative one (you cannot have negative time!).
I hope this helps!