A firecracker shoots up from a hill 150 feet high, with an initial speed of 110 feet per second. Using the formula H(t) = -16t2 + vt + s, approximately how long will it take the firecracker to hit the ground?
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(I take it that the firecracker rises to a maximum height, then falls not only that much but 150 feet more to the bottom of the hill.)
h(t) = -16t² + vt + s.
Here, v = 110 ft/s and s =150 ft, so we can write h(t) = -16t² + 110t + 150.
Take the first derivative, giving
h'(t) = -32t + 110.
When h'(t) = 0, the height is at a maximum:
-32t + 110 = 0, so
32t = 110, so
t = 3.4375 s.
At this time, the firecracker is -16(3.4375)² + 110(3.4375) + 150 = 339 ft high (approximately).
The total fall (to the bottom of the hill) is then 339 + 150 = 489 ft. Put this into the height equation, remembering: (a) that at the beginning of its fall it has zero speed, and (b) we can set the starting point as 0 ft. This makes the landing point be at -489 ft, since we take the positive direction to be upward.
h(t) = -16t² + vt + s becomes
-489 = -16t² + 0t + 0, so
t² = 489/16, so
t = √(489/16) = 5.5 s (approximately). Add this time of falling to the time of rising to max height:
5.5 + 3.4 = 8.9 = 9 s (approximately), answer B.