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bearhunter [10]
3 years ago
15

1. A small light bulb is shining light on a basketball (diameter is 23 cm or 9 inches). The light bulb is 3 m from the closest s

ide of the basketball. Behind the basketball, on the side away from the light bulb, is a wall 4 m from the basketball. Calculate the size (diameter) of the basketball's shadow on the wall
Physics
1 answer:
siniylev [52]3 years ago
4 0

Answer:

The size (diameter) of the basketball's shadow on the wall is approximately 53.38 cm

Explanation:

The given parameters of the basketball are;

The diameter of the basketball = 23 cm (9 inches)

The distance of the light bulb from the closest side of the basketball = 3 m

The distance from the ball to the wall = 4 m

The distance from the light source to the center of the ball, d = 3 m + 0.23/2 m = 3.115 m

The angle the light ray makes with the edge of the ball, θ = arctan(0.115/3.115)

Therefore, the ratio of the shadow width divided by 2 to the distance from the light from the wall = 0.115/3.115

The distance from the light from the wall = 3 m + 4 m + 0.23 m = 7.23 m

Therefore;

((The width of the shadow)/2)/(The distance from the light from the wall) = 0.115/3.115

∴ ((The width of the shadow)/2)/(7.23 m) = 0.115/3.115

((The width of the shadow)/2) = 7.23 m × 0.115/3.115 = 16629/62300 m ≈ 0.2669 m = 26.69 cm

The width (diameter) of the shadow on the wall = 2 × 16629/62300 m ≈ 0.5338 m = 53.38 cm

The size (diameter) of the basketball's shadow on the wall ≈ 53.38 cm

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Answer:

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b. 0.4458 m

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a.

Knowing that

y_o=1.38\ m

We use this formula to compute t

\displaystyle y=y_o-\frac{gt^2}{2}

At the moment when the plate hits the floor y=0

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\displaystyle t=\sqrt{\frac{2(1.38)}{9.8}}

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b.

x=(0.84)(0.5307)

x=0.4458\ m

c.

v_x=0.84\ m/s

v_y=(9.8)(0.5307)

v_y=5.2\ m/s

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