An arrow is shot at 30º above the horizontal.  Its velocity is 50 m/s, and it hits the target.
a.  What is the maximum height the arrow will attain?
b.  The target is at the height from which the arrow was shot.  How far away is it?


We can solve this problem using the principles of projectile motion. The two key components of motion in this problem are the horizontal and vertical components of the velocity. The arrow is shot at an angle of 30º with an initial velocity of 50 m/s.

### Given:
- Initial velocity, \( v_0 = 50 \, \text{m/s} \)
- Angle of projection, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

#### Part a: Maximum Height Attained

The maximum height is attained when the vertical component of the velocity becomes zero. First, we find the vertical component of the initial velocity:

- Vertical component of initial velocity:  
  \[
  v_{0y} = v_0 \sin(\theta) = 50 \times \sin(30^\circ) = 50 \times 0.5 = 25 \, \text{m/s}
  \]

To find the maximum height, we use the following kinematic equation:
\[
v_y^2 = v_{0y}^2 - 2g h_{\text{max}}
\]
At the maximum height, the final vertical velocity \( v_y = 0 \), so:
\[
0 = (25)^2 - 2(9.8) h_{\text{max}}
\]
\[
625 = 19.6 h_{\text{max}}
\]
\[
h_{\text{max}} = \frac{625}{19.6} \approx 31.89 \, \text{m}
\]

Thus, the maximum height attained by the arrow is approximately **31.9 m**.

#### Part b: Range of the Projectile

To find how far away the target is (the range), we need to calculate the total time of flight first. The total time of flight is given by the equation:

\[
t_{\text{total}} = \frac{2 v_{0y}}{g}
\]
Substituting the known values:
\[
t_{\text{total}} = \frac{2 \times 25}{9.8} \approx 5.10 \, \text{seconds}
\]

Now, we can find the horizontal component of the velocity, which remains constant throughout the motion because there is no acceleration in the horizontal direction:

- Horizontal component of initial velocity:  
  \[
  v_{0x} = v_0 \cos(\theta) = 50 \times \cos(30^\circ) = 50 \times 0.866 \approx 43.3 \, \text{m/s}
  \]

The range of the projectile is then:
\[
\text{Range} = v_{0x} \times t_{\text{total}} = 43.3 \times 5.10 \approx 220.83 \, \text{m}
\]

Thus, the arrow will hit the target at a distance of approximately **220.8 m**.

### Summary:
- **Maximum height**: \( 31.9 \, \text{m} \)
- **Range**: \( 220.8 \, \text{m} \)

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Would you like more details on any of these steps or additional questions?

Here are 5 related questions:
1. How would the maximum height change if the angle of projection increased?
2. What is the time to reach the maximum height?
3. How would air resistance affect the range of the projectile?
4. How would the range change if the initial velocity was doubled?
5. How does the angle of projection affect the total flight time?

**Tip:** For maximum range in projectile motion (neglecting air resistance), the optimal angle is 45º.

An arrow is shot at 30º above the horizontal. Its velocity is 50 m/s, and it hits the target. a. What is the maximum height the arrow will attain? b. The target is at the height from which the arrow was shot. How far away is it?

Solve a projectile motion problem where an arrow is shot at an angle of 30º with an initial velocity of 50 m/s. Find the maximum height attained (31.9 m) and the range (220.8 m) using key kinematic formulas. Suitable for students in Grades 9-12 studying physics and trigonometry.

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