Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and the Yahoo Answers website is now in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Please solve (10pts)?
A ball is thrown vertically upward from the top of a building 84 feet tall with an initial velocity of 70 feet per second. The distance S (in feet) of the ball from the ground after T seconds is
S = 96 + 80T - 16t^2
After how many seconds will the ball touch the ground?
Note: the problem is a quadratic equation...
2 Answers
- 1 decade agoFavorite Answer
S = 96 + 80T - 16t^2=0
(-80 +/- √(80^2 -(4)(-16)(96)))/2(-16)=
(-80 +/- √(6400 + 6144))/-32=
(-80 +/- √12544)/-32=
(-80 +/- 112)/-32=
t = 6 or -1
but you can't have negative time so t = 6
- 1 decade ago
The answer to 0 = 96 + 80T - 16t^2
Is t = 6, though the equation you gave doesn't seem to match the problem description.
The reason I give for this is at t = 0, the ball should be at the top of the building, S = 84 ft, not S = 96 ft as the equation gives (unless the person who threw it is close to 12 ft tall). You might want to double check your info.
