You buy a plastic dart gun, and being a clever physics student you decide to do a quick calculation to find its maximum horizontal range. You shoot the gun straight up, and it takes 4.4 s
s
for the dart to land back at the barrel. What is the maximum possible horizontal range that could be achieved by your gun, assuming the dart lands at the same level it started?
Express your answer using two significant figures and include the appropriate units.


To find the maximum possible horizontal range of the dart gun, we can follow these steps:

1. **Determine the initial vertical velocity:**
   The total time it takes for the dart to travel up and back down is 4.4 seconds. Since this is a projectile motion problem, the time to reach the maximum height is half of the total time. Therefore, the time to reach the maximum height is:

   \[
   t_{\text{up}} = \frac{4.4}{2} = 2.2 \, \text{seconds}
   \]

   Using the equation for vertical motion under constant acceleration due to gravity (\( g = 9.8 \, \text{m/s}^2 \)):

   \[
   v_{0y} = g \cdot t_{\text{up}}
   \]

   where \( v_{0y} \) is the initial vertical velocity. Substituting the known values:

   \[
   v_{0y} = 9.8 \, \text{m/s}^2 \times 2.2 \, \text{seconds} = 21.56 \, \text{m/s}
   \]

2. **Determine the maximum horizontal range:**
   The maximum horizontal range is achieved when the dart is launched at a \( 45^\circ \) angle. At this angle, the horizontal and vertical components of the initial velocity are equal.

   The total initial velocity \( v_0 \) is related to the vertical component \( v_{0y} \) by the following equation (since at a \( 45^\circ \) angle, \( v_{0x} = v_{0y} \)):

   \[
   v_0 = \frac{v_{0y}}{\sin 45^\circ} = \frac{21.56}{\frac{\sqrt{2}}{2}} = 30.48 \, \text{m/s}
   \]

3. **Calculate the range:**
   The range \( R \) of a projectile is given by the equation:

   \[
   R = \frac{v_0^2 \sin 2\theta}{g}
   \]

   For a \( 45^\circ \) launch angle, \( \sin 2\theta = \sin 90^\circ = 1 \). Substituting the values:

   \[
   R = \frac{30.48^2}{9.8} = \frac{929.06}{9.8} = 94.8 \, \text{m}
   \]

So, the maximum horizontal range is:

\[
\boxed{95 \, \text{m}}
\]

Let me know if you want further details or have any questions!

---

Here are 5 relative questions for further practice:
1. What would the range be if the gun was shot at an angle of 30 degrees?
2. How does air resistance affect the maximum range of the dart?
3. What is the maximum height the dart reaches when shot vertically upward?
4. How would the range change if gravity on Earth was different, say on the Moon?
5. What is the effect of the mass of the dart on the range?

**Tip:** For projectiles, a 45° launch angle provides the maximum horizontal range in a vacuum.

You buy a plastic dart gun, and being a clever physics student you decide to do a quick calculation to find its maximum horizontal range. You shoot the gun straight up, and it takes 4.4 s for the dart to land back at the barrel. What is the maximum possible horizontal range that could be achieved by your gun, assuming the dart lands at the same level it started? Express your answer using two significant figures and include the appropriate units.

Calculate the maximum horizontal range of a dart gun using projectile motion principles. This detailed solution walks you through determining the initial velocity, applying trigonometry, and using the projectile range formula. Ideal for Grades 10-12 students studying physics.

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