SOLUTION: When the shot whose path is released at an angle of 65°, its height, g(x), in feet, can be modeled by: g(x) = —0.04x2 + 2.1x + 6.1 where x is the shot’s horizontal distance, in fee

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Question 946580: When the shot whose path is released at an angle of 65°, its height, g(x), in feet, can be modeled by: g(x) = —0.04x2 + 2.1x + 6.1 where x is the shot’s horizontal distance, in feet, from its point of release. Use this model to solve the following: (a) What is the maximum height, to the nearest tenth of a foot, of the shot and how far from its point of release does this occur? (b) What is the shot’s maximum horizontal distance, to the nearest tenth of a foot, or the distance of the throw? (c) from what height was the shot released?
Answer by josmiceli(19441)   (Show Source): You can put this solution on YOUR website!

This is in the form:
where



-------------
(a)
The component of the maximum
height is :




--------------------



To the nearest 10th, the maximum height
is 33.7 ft
The maximum height is reached 26.25 ft
from the point of release
-----------------------
(b)


Use the quadratic formula










The maximum horizontal distance is:
55.3 ft to the nearest tenth
-------------------------
(c)
To find height from which shot is released, se the
horizontal distance, to


The shot was released from a height of 6.1 ft
---------------------------
Here's a plot:




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