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Nezavi [6.7K]
3 years ago
6

An object is launched upward at 64 ft/sec from a platform that is 80 feet high. What is the object's maximum height if the equat

ion of height (h) in terms of time (t) of the object is given by h(t) = -16t2 + 64t + 80?
Mathematics
1 answer:
daser333 [38]3 years ago
4 0
The object's maximum height is the same as the y-coordinate of the vertex.To find that, first, finds the x- coordinate(t).You can find it by:
first identifying a,b and c.Multiply the b by negative 1 or simply change its sign and divide it by 2a.Afterward, plug the answer in the equation to find y, your answer.
(ax+bx²+c=-16t²+64t+80). Therefore, a=-6, b= 64 and c=80
-64/2(-16)=-64/-32=2
-16(2)²+64(2)+80=144 ft
So the maximum height of the object, when launched 64ft/sec from a platform that is 80 ft high, is 144 ft
Hope this helps(there are other methods)
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