An object is launched at 19.6 m/s from a height of 58.8 m. The equation for the height (h) in terms of time (t) is given by h(t)
= -4.9t2 +19.6t + 58.8. What is the object's maximum height?
1 answer:
H(t) = Ho +Vot - gt^2/2
Vo = 19.6 m/s
Ho = 58.8 m
g = 9.8 m/s^2
H(t) = 58.8 + 19.6t -9.8t^2/2 = 58.8 + 19.6t - 4.9t^2
Maximun height is at the vertex of the parabole
To find the vertex, first find the roots.
58.8 + 19.6t - 4.9t^2 = 0
Divide by 4.9
12 + 4t - t^2 = 0
Change sign and reorder
t^2 - 4t -12 = 0
Factor
(t - 6)(t + 2) =0 ==> t = 6 and t = -2.
The vertex is in the mid point between both roots
Find H(t) for: t = [6 - 2]/2 =4/2 = 2
Find H(t) for t = 2
H(6) = 58.8 + 19.6(2) - 4.9(2)^2 = 78.4
Answer: the maximum height is 78.4 m
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