An object is thrown from the origin of a coordinate system with the x-axis along the groundand the y-axis vertical. Its path, or tragectory, is given by the equation y=359x-13x^2. Find the objects maximum height.
Hello, epetrik!
Here is a non-Calculus solution . . .An object is thrown from the origin of a coordinate system
with the x-axis along the ground and the y-axis vertical.
Its path, or trajectory, is given by the equation: .$\displaystyle y\:=\:359x-13x^2$
Find the object's maximum height.
The path is a down-opening parabola; its maximum is at its vertex.
The vertex of the parabola, $\displaystyle y \:=\:ax^2 + bx + c$, is at: .$\displaystyle x \:=\:\frac{\text{-}b}{2a}$
We have: .$\displaystyle a = \text{-}13,\;b = 359$
. . The vertex is at: .$\displaystyle x \:=\:\frac{\text{-}359}{2(\text{-}13)} \:=\:\frac{359}{26}$
Then: .$\displaystyle y \:=\: 359\left(\frac{359}{26}\right) - 13\left(\frac{359}{26}\right)^2 \:=\:\frac{128,\!881}{52} $
The maximum height is about: $\displaystyle 2478.48$ units.