Let f(x)=3x^2+6x find f(2)A.-4B. 0C. 4D. 8
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Answer:
a. s = 64t - 16t²
a. maximum height = 64 ft
b. time it took to reach maximum height = 2 s
c. total time it spent in the air = 4 s
Step-by-step explanation:
a. Using Newton's third equation of motion under gravity, s = ut - 1/2gt² where s = vertical height of rocket, t = time, u = initial velocity of rocket = 64 ft/s and g = acceleration due to gravity = 32 ft/s².
So, s = 64t - 1/2(32)t² = 64t - 16t²
The equation is thus s = 64t - 16t²
b. The maximum height is obtained when ds/dt = 0
ds/dt = d[64t - 16t²]/dt = 64 - 32t
64 - 32t = 0
32t = 64
t = 64/32 = 2 s
Substituting t = 2 into s, we have
s = 64(2) - 16(2)² = 128 - 64 = 64 ft
c. The amount tot time it took to reach maximum height is 2 s
d. The amount of time it was in the air is obtained when s = 0
s = 0 ⇒
64t - 16t² = 0
16t(4 - t) = 0
16t = 0 or 4 - t = 0
t = 0 or t = 4
So the rocket spends 4 s in the air.
The values of the different variable for a, b,c and d are indicated in the graph.
The point (2,64) indicated the time for maximum height 2 s and maximum height 64 ft.
The point (4,0) indicates the total time spent in the air 4 s
