Respuesta :
The change of the distance travelled by the ball when the angle of launch
changes can be found by linear approximation.
- The change in the distance, Δs, when the angle changes by Δθ in radians is approximately 0.211·Δθ.
Reasons:
The distance of the player from the basket = 18.6 ft.
Height from which the jump shot is launched = 10 ft.
Angle of the jump shot = 36°
The initial velocity = 25 ft./s
According to Newton's Law of motion, the distance an object travels when
released at an angle θ, with an initial velocity, v, is given by the following
horizontal range formula equation;
[tex]s = \dfrac{1}{32} \cdot v^2 \cdot sin(2 \cdot \theta)[/tex]
[tex]\dfrac{ds}{d\theta} =\dfrac{d}{d\theta } \left( \dfrac{1}{32} \cdot v^2 \cdot sin(2 \cdot \theta) \right)= \dfrac{v^2 \cdot cos(2 \cdot \theta)}{16}[/tex]
By linear approximation, we have;
[tex]\Delta s = \dfrac{ds}{d\theta} \times \Delta \theta[/tex]
Which gives;
[tex]\Delta s =\dfrac{25^2 \cdot cos(2 \times 36^{\circ})}{16}\times \Delta \theta \approx 12.07 \times \Delta \theta[/tex]
360° = 2·π radians
[tex]12.07 ^{\circ} = \dfrac{2 \cdot \pi}{360} \times 12.07 \approx 0.211 \ radians[/tex]
Δs ≈ 0.211·Δθ
Therefore;
The change in the distance, Δs, when the angle changes by Δθ is 0.211·Δθ.
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