Answer: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:
