A baseball slugger hits a pitch and watches the ball fly into the bleachers for a home run, landing h = 9.5 m higher than it was struck. When visiting with the fan that caught the ball, he learned the ball was moving with final velocity vf = 30.15 m/s at an angle θf = 29° below horizontal when caught. Assume the ball encountered no air resistance, and use a Cartesian coordinate system with the origin located at the ball's initial position.

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davidlcastiblanco davidlcastiblanco · Answer 1 · 2019-10-02T17:53:03+02:00

Answer:

The ball's position can get of the initial angle when leave the bat and velocity magnitud.

θ = 35,1

v = 36,5 [tex]\frac{m}{s}[/tex]

Explanation:

θf= 29

h= 9,5 m

vf= 30,15 [tex]\frac{m}{s}[/tex]

[tex]V_{ox} = V_{f} * cos (θ)[/tex]

[tex]V_{ox} = 30, 15 * cos ( 29 )[/tex]

[tex]V_{ox} = 26,36 \frac{m}{s}[/tex]

Now as you know the vector in 'y' can be related with Tan function

[tex]Tan (θ) = \frac{V_{y} }{V_{ox} }[/tex]

[tex]V_{y} = Tan (29) * 26,36 \frac{m}{s}[/tex]

[tex]V_{y} = 14,617 \frac{m}{s}[/tex]

So, the initial velocity in 'y' can be resolved:

[tex]V_{oy} ^{2}= V_{y} ^{2} + 2* a * h[/tex]

[tex]V_{y} ^{2} = 14,617^{2} +2 * 9,8 \frac{m}{s^{2}} *9,5 m[/tex]

[tex]V_{oy} = \sqrt{399,8569828}[/tex]

[tex]V_{oy} =19,99642525 \frac{m}{s} [/tex]

Finally the velocity is going to be:

[tex]V= \sqrt{V_{ox} ^{2} +V_{oy} ^{2} }[/tex]

[tex]V= \sqrt{26,36 ^{2} +19,99 ^{2} }[/tex]

[tex]V= 33,08 \frac{m}{s}[/tex]

θ[tex]= Tan^{-1} (\frac{19,99}{26,36})[/tex]

θ[tex]=37,188[/tex]