Answer to Question #236150 in Calculus for mika

Question #236150

An object thrown from a height of 2 m above the ground follows a parabolic path

until the object falls to the ground; see Figure. If the object reaches a maximum

height (measured from the ground) of 7 m after travelling a horizontal distance of 4

m, determine the horizontal distance between the object's initial and final positions.


1
Expert's answer
2021-09-12T23:52:36-0400

The motion can be broken into horizontal and vertical motions


"vertical: y(t)=h_0+v_{0y}t-\\dfrac{gt^2}{2}"

"horizontal : x(t)=v_{0x}t"

The highest point in any trajectory, called the apex, is reached when


"v_y(t_1)=v_{0y}-gt_1=0"

Then


"y(t_1)=h_{max}=h_0+v_{0y}(\\dfrac{v_{0y}}{g})-\\dfrac{g}{2}(\\dfrac{v_{0y}}{g})^2=h_0+\\dfrac{v_{0y}^2}{2g}"

"=h_0+\\dfrac{v_{0y}^2}{2g}"

"v_{0y}=\\sqrt{2g(h_{max}-h_0)}"

"v_{0y}=\\sqrt{2(9.8\\ m\/s^2)(7\\ m-2\\ m)}=7\\sqrt{2}\\ m\/s"

"x(t_1)=v_{0x}t_1=\\dfrac{v_{0x}v_{0y}}{g}"

"v_{0x}=\\dfrac{4\\ m(9.8\\ m\/s^2)}{7\\sqrt{2}\\ m\/s}=2.8\\sqrt{2}\\ m\/s"


The object will fall to the ground in time "t_2"

"y(t_2)=h_0+v_{0y}t_2-\\dfrac{gt_2^2}{2}=0"

"2+7\\sqrt{2}t_2-\\dfrac{9.8t_2^2}{2}=0, t_2>0"

"4.9t_2^2-7\\sqrt{2}t_2-2=0"

"D=(-7\\sqrt{2})^2-4(4.9)(-2)=137.2"

"t_2=\\dfrac{7\\sqrt{2}\\pm\\sqrt{137.2}}{2(4.9)}=\\dfrac{\\sqrt{2}\\pm\\sqrt{2.8}}{1.4}"


Since "t_2>0," then we take


"t_2=\\dfrac{\\sqrt{2}+\\sqrt{2.8}}{1.4}"


The horizontal distance between the object's initial and final positions is


"d=x(t_2)-x(0)=v_{0x}t_2-0"

"d=2.8\\sqrt{2}(\\dfrac{\\sqrt{2}+\\sqrt{2.8}}{1.4})=4(1+\\sqrt{1.4}) (m)"




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