Question

if a firework shell has a 3.5 in. diameter, how high would you expect it to travel before bursting? enter your answer in the box. \boxed{} m.

Explanation:

Step1: Recall the relationship (assuming a linear relationship, e.g., from a model or formula: height (m) = k diameter (in), where k is a conversion factor considering physics of projectile motion, or a standard model for firework shells. A common approximation: for firework shells, height (in feet) can be related, but we need meters. First, convert diameter to meters: 3.5 in = 3.5 0.0254 m ≈ 0.0889 m. But maybe a better approach: typical formula (empirical) for firework shell height: \( h = 100 \times d \) (where d is diameter in inches, h in meters? Wait, no, maybe feet. Wait, let's check: 1 inch = 0.0254 m. Let's assume a model: height (m) = (diameter in inches) 25.4 cm/inch some factor. Wait, maybe the standard is that for a firework shell, the height (in feet) is about 100 times the diameter (in inches), then convert to meters. So 3.5 in diameter: height in feet = 3.5 100 = 350 feet. Convert to meters: 350 ft 0.3048 m/ft ≈ 106.68 m. Wait, but maybe the formula is different. Alternatively, maybe the problem assumes a linear relationship from a graph or prior data. Let's proceed with the conversion:

First, convert 3.5 inches to meters: \( 3.5 \times 0.0254 = 0.0889 \) m. But that's not the height. Wait, maybe the formula is \( h = 30 \times d \) (d in inches, h in meters)? No, that doesn't make sense. Wait, perhaps the correct approach is that in physics, the height of a projectile depends on initial velocity, but for fireworks, the initial velocity is related to the shell size. A common rule of thumb: the height (in meters) of a firework shell is approximately 30 times the diameter (in inches). Wait, 3.5 * 30 = 105 m. Alternatively, using the formula for projectile motion: \( h = \frac{v_0^2 \sin^2 \theta}{2g} \). For a firework, θ is 90 degrees (vertical), so \( h = \frac{v_0^2}{2g} \). The initial velocity v0 is related to the shell's mass and charge, but for a 3.5 in shell, typical v0 is about 100 m/s? No, that's too fast. Wait, 350 feet is about 106 meters, as before. Let's check: 3.5 inches diameter, height ~ 350 feet = 106.68 meters. So we'll use that.

Step2: Calculate the height. Using the rule of thumb: height (m) = (3.5 in 100 ft/in) 0.3048 m/ft = 350 0.3048 ≈ 106.68 m, which rounds to approximately 107 m, but maybe the problem expects a simpler calculation. Wait, maybe the formula is h = 30 d, where d is in inches, so 3.5 * 30 = 105 m. Alternatively, maybe the answer is around 106 or 107 meters. Let's confirm:

3.5 inches diameter. Convert to meters: 3.5 0.0254 = 0.0889 m (diameter), but height is not that. Wait, I think the key is that the problem is likely using a linear model where height (m) = 30 diameter (in), so 3.5 30 = 105 m, or 100 3.5 * 0.3048 ≈ 106.68 m. So the expected answer is approximately 107 m, but let's check the calculation again.

Wait, maybe the correct formula is that for a firework shell, the height (in meters) is 30 times the diameter (in inches). So 3.5 30 = 105 m. Or using the feet to meters: 3.5 100 = 350 feet, 350 * 0.3048 = 106.68 m, so approximately 107 m.

Answer:

\boxed{107} (or 106.68, rounded to 107)