You should fly at an angle of [tex]14.0^{\circ}[/tex]
Explanation:
The motion of the ball is a free fall motion, so the vertical displacement of the ball (in negative directions) is given by
[tex]y=\frac{1}{2}gt^2[/tex]
where
[tex]g=9.8 m/s^2[/tex] is the acceleration due to gravity
t is the time
The horizontal distance you have to cover to catch the ball is
[tex]x = 30 m=d[/tex]
The vertical and horizontal displacement form a right triangle, so that we can write
[tex]tan \theta = \frac{y}{x}[/tex]
where [tex]\theta[/tex] is the direction of your motion. Substituting,
[tex]tan \theta = \frac{gt^2}{2d}[/tex] (1)
We also know that the horizontal position of the person at time t is
[tex]x(t) = v cos \theta t[/tex]
And so, for the time t at which the person catches the ball,
[tex]d=v cos \theta t\\t=\frac{d}{v cos \theta}\\t^2=\frac{d^2}{v^2 cos^2 \theta}[/tex]
Substituting this expression into (1),
[tex]tan \theta = \frac{g}{2d} \frac{d^2}{v^2 cos^2 \theta}\\2 tan \theta cos^2 \theta = \frac{gd}{2v^2}[/tex]
And using
[tex]tan \theta = \frac{sin \theta}{cos \theta}, 2sin \theta cos \theta = sin(2\theta)[/tex]
we find:
[tex]sin (2\theta) = \frac{gd}{v^2}=\frac{(9.8)(30)}{25^2}=0.470\\2\theta = sin^{-1}(0.470)=28.0^{\circ}\\\theta=\frac{28}{2}=14.0^{\circ}[/tex]
Learn more about free fall motion:
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