The equation h=-16t^2 + 112t gives the height of an arrow, shot upward from the ground with an initial velocity of 112 ft/s, where t is the time after the arrow leaves the ground. Find the time it takes for the arrow to reach a height of 180 ft.
2006-10-11 09:20:50 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
Set h=180 and solve for t.
That gives you the equation
-16t^2 + 112t = 180, or, after rearranging,
16t^2 - 112t + 180 = 0.
Divide by 4:
4t^2 - 28t + 45 = 0
Factor:
(2t-5)(2t -9) = 0
so t=5/2, or t=9/2. Choose the lower value.
2006-10-11 09:25:01 · answer #1 · answered by James L 5 · 0⤊ 0⤋
This is a simple quadratic equation:
16t^2 - 112t +180 = 0
t = (-(-112) +/- sqroot(112^2 + 4*16*180))/(2*16)
2006-10-11 16:26:48 · answer #2 · answered by T 5 · 0⤊ 0⤋
180==-16t^2 + 112t
-16t^2+112t-180=0
16t^2-112t+180=0
4t^2-28t+45=0
t=(28+/-sqrt(28^2-4*4*45))/8=(28+/-sqrt(784-720))/8
t=(28+/-sqrt(64))/8=(28+/-8)/8
t=(28+8)/8=4.5 sec
t=(28-8)/8=2.5 sec
it takes 2.5 sec for the arrow to reach 180 ft on the way up & 4.5 sec to passthrough 180ft on the way down.
2006-10-14 22:52:39 · answer #3 · answered by yupchagee 7 · 0⤊ 0⤋
180 = -16t^2 + 112t
so
45 = -4t^2 + 28t so 4t^2 - 28t + 45 = 0
solve with quadratic solution
2006-10-11 16:26:23 · answer #4 · answered by ? 7 · 0⤊ 0⤋