Answer:
the object is in the air on the time interval (0.24 sec, 6.51 sec)
Step-by-step explanation:
The object is 'in the air' for all t such that h> 0. We need to find the roots of h = -16t^2 + 108t - 25 = 0. From the graph we see that both t values are positive. Once we find them, we subtract the smaller t from the larger t, which results in the length of time the object is in the air.
Use the quadratic formula to find the roots of h(t). The coefficients of t are {-16, 108, -25}, and so the discriminant b^2 - 4ac is
108² - 4(-16)(-25) = 11664 - 1600 = 10064, whose square root is 100.32.
Then the quadratic formula x = (-b ± √[b² - 4ac)/(2a) becomes
-108 ± 100.32 108 ± 100.32
t = ---------------------- = --------------------- = 3.375 ± 3.135
2(-16) 32
or t = 6.51 or t = 0.24 (both times expressed in seconds).
So, again, the object is in the air on the time interval (0.24 sec, 6.51 sec)
