9514 1404 393
Answer:
a) 2.038 seconds
b) 5.918 meters
c) 1.076 seconds
Step-by-step explanation:
For the purpose of answering these questions, it is convenient to put the given equation into vertex form.
h = -4.9t² +9.2t +1.6
= -4.9(t² -(9.2/4.9)t) +1.6
= -4.9(t² -(9.2/4.9)t +(4.6/4.9)²) +1.6 +4.9(4.6/4.9)²
= -4.9(t -46/49)² +290/49
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a) To find h = 0, we solve ...
0 = -4.9(t -46/49)² +290/49
290/240.1 = (t -46/49)² . . . . subtract 290/49 and divide by -4.9
√(2900/2401) +46/49 = t ≈ 2.0378 . . . . seconds
The ball takes about 2.038 seconds to fall to the ground.
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b) The maximum height is the h value at the vertex of the function. It is the value of h when the squared term is zero:
290/49 m ≈ 5.918 m
The maximum height of the ball is about 5.918 m.
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c) We want to find t for h ≥ 4.5.
h ≥ 4.5
-4.9(t -46/49)² +290/49 ≥ 4.5
Subtracting 290/49 and dividing by -4.9, we have ...
(t -46/49)² ≥ 695/2401
Taking the square root, and adding 46/49, we find the time interval to be ...
-√(695/2401) +46/49 ≤ t ≤ √(695/2401) +46/49
The difference between the interval end points is the time above 4.5 meters. That difference is ...
2√(695/2401) ≈ 1.076 . . . . seconds
The ball is at or above 4.5 meters for about 1.076 seconds.
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I like a graphing calculator for its ability to answer these questions quickly and easily. The essentials for answering this question involve typing a couple of equations and highlighting a few points on the graph.
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<em>Additional comment</em>
I have a preference for "exact" answers where possible, so have used fractions, rather than their rounded decimal equivalents. The calculator I use deals with these fairly nicely. Unfortunately, the mess of numbers can tend to obscure the working.
"Vertex form" for a quadratic is ...
y = a(x -h)² +k . . . . where the vertex is (h, k) and 'a' is a vertical scale factor.
In the above, we have 'a' = -4.9, and (h, k) = (46/49, 290/49) ≈ (0.939, 5.918)
