The function y= -16t^2 + 276 models the height (y) in feet of a stone (t) seconds after it is dropped from the edge of a vertical cliff. How long will it take the stone to hit the ground? Round to the nearest hundredth of a second.
2006-12-01 05:37:06 · 5 answers · asked by Reese's Pieces 2 in Science & Mathematics ➔ Mathematics
You want to know at what time y=0 since that represents when the rock is on the ground.
0=-16t^2+276
16t^2=276
t^2=17.25
t=sqrt(17.25)=4.1533
Ans: 4.15 seconds
2006-12-01 05:40:09 · answer #1 · answered by hobbes84k 3 · 0⤊ 0⤋
When the stone hits the ground, what will its height be? It will be zero, right? The question is basically asking at what time (t) will the height above ground (y) be zero.
So set the equation to zero and solve for t:
0 = -16t² + 276
Add 16t² to both sides:
16t² = 276
Divide both sides by 16:
t² = 276 / 16
Reduce to lowest terms:
t² = 69 / 4
Take the square root of both sides (technically there is a negative square root, but time(t) in this case will always be positive, so you can ignore the negative root).
t = sqrt(69 / 4)
Simplify:
t = sqrt(69) / 2
Calculate and round to the nearest hundredth of a second:
t â 4.15 seconds
2006-12-01 13:40:22 · answer #2 · answered by Puzzling 7 · 0⤊ 0⤋
When the stone hits the ground the height will be 0
Hence 0 = -16t^2 + 276
16t^2 = 276
t^2 = 17.25
t = 4.15 to nearest hundreth of second
2006-12-01 13:40:52 · answer #3 · answered by Anonymous · 0⤊ 0⤋
The ground is height y=0, so you just need to solve -16t^2 + 276 = 0. Good luck!
2006-12-01 13:40:13 · answer #4 · answered by Anonymous · 0⤊ 0⤋
set y = 0 and solve for t....
0 = -16t^2 +276
divide both sides by -16 and subtract the 276/-16
t^2 = 17.25
t = 4.15 or 4 secs...
2006-12-01 13:41:08 · answer #5 · answered by beatdawookie 2 · 0⤊ 0⤋