the function h defined by h(t) = (49+4.9t) (10-t) models the height, in meters, of an object, in meters of an object t seconds after it is dropped from a helicopter. Find the approximate time when the object hits the the ground ​

h(t)=(49+4.9t)(10-t)

Notice that h goes from the initial height of the helicopter down to zero.

t goes from 0 ("after it is dropped") up to the maximum t (when the object hits the ground)

Actually, it is easier to find h first (part b), so let's do that first:

At t=0,

h = (49)(10)

h = 490 m

O.K., now find maximum t (part a).

Well, at maximum t, we know that h = 0 ("hits the ground")

h(0) = (49+4.9t)(10-t)

0 = (49+4.9t)(10-t)

0 = 490 + 490t - 49t - 4.9t (remember F-I-O-L)

0 = 490 - 436.1t

-490 = -436.1t

1.12 ≅ t

Now, measuring one tenth or one hundredth of one hundredth of a second requires very special instruments (not just a stopwatch), so you should say that the object hit the ground at t=1 second.

Let's check to see how close we are:

h(1) =(49+4.9(1))(10-1)

h(1) = (49 + 4.9)(9)

h(1) = 485.1 meters

This check gave 485.1 meters instead of 490 meters. For 490 meters, being off by less than 5 meters is close enough to be called a "good estimate."

Step-by-step explanation:

it would rotate about 180 degrees.