Question

[9.06] 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?

Answer 1

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 2

H(t) = -16t^2 + vt + s
0 = -16t^2 + 110t + 150
8t^2 - 55t - 75 = 0
t = 8.04 seconds.