Question

I need this ASAP:
a flare is fired from the bottom of a gorge is visible only when the flare is above the rim. if it is fired with an initial velocity of 96ft/sec, and the gorge is 128ft deep, during what interval can the flare be seen?\[h=-16t^{2}+v_{o}t+h_{o}\]

Answer 1

@Xishem

Answer 2

initial height is 0 im assuming
you have initial velocity
substitute those in and find t when h = 128

Answer 3

substitute the values where h = 128 and v= 96 h0= 0
u get a quadratic equation solve it and find the time

Answer 4

your answer would lie between the 2 t values you calculate

Answer 5

128 = -16t^2 +96t +0
16t^2 -96t + 128 = 0

use the quadratic formula and find the values, u will get two answers, subtract the small value from the large value

Answer 6

\[128=16t^{2}+96t+0\]\[0=16t^{2}+96t-128\]\[0=t^{2}+6t-8\]\[0=(t+4)(t-2)\]\[t=-4 ~~~or~~~ t=2\]

Answer 7

@thushananth01 ??

Answer 8

That is correct, the flare is seen from t = 2 seconds to t = 4 seconds

Answer 9

but the 4 is negative...?

Answer 10

it shouldn't be, it's positive

Answer 11

\[\Large t = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]


\[\Large t = \frac{-(-96)\pm\sqrt{(-96)^2-4(16)(128)}}{2(16)}\]


\[\Large t = \frac{96\pm\sqrt{9216-(8192)}}{32}\]


\[\Large t = \frac{96\pm\sqrt{1024}}{32}\]


\[\Large t = \frac{96+\sqrt{1024}}{32} \ \text{or} \ t = \frac{96-\sqrt{1024}}{32}\]


\[\Large t = \frac{96+32}{32} \ \text{or} \ t = \frac{96-32}{32}\]


\[\Large t = \frac{128}{32} \ \text{or} \ t = \frac{64}{32}\]


\[\Large t = 4 \ \text{or} \ t = 2\]

Answer 12

Sorry for some reason, I didn't see that negative...not sure how you got -4, but it should be positive 4

Answer 13

YES use the quadratic formula, my friend..u can find time directly..:)

Answer 14

k thanks!!

Answer 15

no probs:)