An archery enthusiast was testing his new bow. He shot an arrow straight up from the top of a building 48 feet high with an initial velocity of 32 feet per second. The height of the arrow (h measured in feet) after t seconds in flight is described by the equation
h = 48 + 32 t - 16 t**2.
Determine the number of seconds after the arrow is shot that it will strike the ground.
t = seconds
2007-09-02 11:13:29 · 5 answers · asked by lauren m 1 in Science & Mathematics ➔ Mathematics
1 . Set h = 0.
2. Write equation:
0 = 48 + 32t -16t^2
3. Divide through by -16:
0 = -3 -2t + t^2
4. Factor:
(t-3)(t+1)
5. Look at the answers:
t= 3
t= -1.
6, Through out the one that doesn't make sense (-1).
7. Write your answer: t = 3.
.
I should note that if the arrow were shot straight up, it would hit the roof and not the ground. It doesn't sound like that's what your teacher wants, though.
2007-09-02 11:18:47 · answer #1 · answered by Anonymous · 0⤊ 0⤋
If he is on top of a building and shoots the arrow straight up, why will the arrow hit the ground and not the top of the building or even the top of his/her head?
48 +32t -16t^2 = 0
3 + 2t -t^2 = 0
(3-t)(1+t) = 0
t = 3 seconds
2007-09-02 18:24:43 · answer #2 · answered by ironduke8159 7 · 0⤊ 0⤋
at the end when you say 16t**2 do you mean 16 time t squared? If so arrow struck ground when t=3
2007-09-02 18:23:19 · answer #3 · answered by cj 2 · 0⤊ 0⤋
Make h=0.
0 = 48 + 32t - 16t^2
0 = -16t^2 + 32t + 48
Divide by -16
0 = t^2 - 2t - 3
Factor...
(t-3)(t+1) = 0
t=3 or t=-1
Time can't be negative, so t=3!
2007-09-02 18:21:49 · answer #4 · answered by JO 3 · 0⤊ 0⤋
You failed to say when both trains are leaving the stations!
2007-09-02 18:16:54 · answer #5 · answered by Anonymous · 1⤊ 0⤋