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The question is an illustration of related rates.
The rate of change between you and the ball is 0.01 rad per second
I added an attachment to illustrate the given parameters.
The representations on the attachment are:
[tex]\mathbf{x = 100\ m}[/tex]
[tex]\mathbf{\frac{dy}{dt} = 2\ ms^{-1}}[/tex] ---- the rate
[tex]\mathbf{\theta = \frac{\pi}{4}}[/tex]
First, we calculate the vertical distance (y) using tangent ratio
[tex]\mathbf{\tan(\theta) = \frac{y}{x}}[/tex]
Substitute 100 for x
[tex]\mathbf{y = 100\tan(\theta) }[/tex]
[tex]\mathbf{\tan(\theta) = \frac{y}{100}}[/tex]
Differentiate both sides with respect to time (t)
[tex]\mathbf{ \sec^2(\theta) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot \frac{dy}{dt}}[/tex]
Substitute values for the rates and [tex]\mathbf{\theta }[/tex]
[tex]\mathbf{ \sec^2(\pi/4) \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}[/tex]
This gives
[tex]\mathbf{ (\sqrt 2)^2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}[/tex]
[tex]\mathbf{ 2 \cdot \frac{d\theta}{dt} = \frac{1}{100} \cdot 2}[/tex]
Divide both sides by 2
[tex]\mathbf{ \frac{d\theta}{dt} = \frac{1}{100} }[/tex]
[tex]\mathbf{ \frac{d\theta}{dt} = 0.01 }[/tex]
Hence, the rate of change between you and the ball is 0.01 rad per second
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