To find the projection angle ($\theta$) needed for the arrow to strike a target at the same elevation, we can use the following kinematic equations for projectile motion. Since the initial and final vertical positions are the same, we can apply this formula for range:

$R = \frac{v_0^2 \sin(2\theta)}{g}$

Where:

• $R = 37$ m (range or distance to the target),
• $v_0 = 26$ m/s (initial speed),
• $g = 9.8$ m/s² (acceleration due to gravity),
• $\theta$ is the projection angle.

We need to solve for $\theta$:

$\sin(2\theta) = \frac{R \cdot g}{v_0^2}$

First, calculate $\sin(2\theta)$:

$\sin(2\theta) = \frac{37 \cdot 9.8}{26^2}$

Now let's compute this.We have:

$\sin(2\theta) = 0.5364$

Now, to find $2\theta$, we take the inverse sine (arcsin) of this value:

$2\theta = \sin^{-1}(0.5364)$

Next, divide by 2 to find $\theta$:

$\theta = \frac{\sin^{-1}(0.5364)}{2}$

Let me compute that for you.The projection angle $\theta$ should be approximately $16.22^\circ$ for the arrow to strike the target 37 meters away at the same elevation.

Would you like more details or further explanations? Here are five related questions to expand your understanding:

1. How would the angle change if the initial speed were increased?
2. What happens to the range if we increase the projection angle beyond $45^\circ$?
3. How does air resistance affect the ideal projection angle?
4. Can you calculate the maximum height reached by the arrow?
5. How long does the arrow stay in the air for this projection angle?

Tip: For maximum range on level ground, the ideal projection angle is $45^\circ$.