Question

A projectile is fired straight up from ground level with an initial velocity of 112 ft/s. Its height, h, above the ground after t seconds is given by h = –16t2 + 112t. What is the interval of time during which the projectile's height exceeds 192 feet? please help me!

Answer 1

So you set h to 192, then set it up as an inequality
192ft<(112ft/s)t−(16ft/s2)t2

So the 112t - 16t^2 has to be greater than 192 ft. Then you come up with the polynomial

−(16ft/s2)t2+(112ft/s)t−192ft>0

to make arithmetic simpler, that you can reduce to
−(1ft/s2)t2+(7ft/s)t−(12ft)=0

of the form
at2+bt+c>0


and solve for t using the quadratic formula - time will be greater than the smaller root, and less than the larger root

t<−b−b2−4ac−−−−−−−√2a;t>−b+b2−4ac−−−−−−−√2a

a=−1b=+7c=−12

Answer 2

h=-16t^2+112

Answer 3

3 < t < 4
t < 4
t > 4
3 > t > 4
these are the choices you get

Answer 4

what was the answer???