If an arrow is shot at 30 degrees with the horizontal and has a velocity of 49 m/s how high will the arrow go and what horizontal distance will the arrow travel?

To solve this, we can break the initial velocity of the arrow into horizontal and vertical components and use kinematic equations. Here’s a step-by-step solution.

### Given Data
- Initial velocity, \( v_0 = 49 \, \text{m/s} \)
- Launch angle, \( \theta = 30^\circ \)
- Acceleration due to gravity, \( g = 9.8 \, \text{m/s}^2 \)

### Step 1: Resolve the Initial Velocity
The velocity can be split into:
1. **Horizontal Component**: \( v_{0x} = v_0 \cos \theta \)
2. **Vertical Component**: \( v_{0y} = v_0 \sin \theta \)

Calculating these:
\[
v_{0x} = 49 \cos(30^\circ) = 49 \times \frac{\sqrt{3}}{2} \approx 42.44 \, \text{m/s}
\]
\[
v_{0y} = 49 \sin(30^\circ) = 49 \times 0.5 = 24.5 \, \text{m/s}
\]

### Step 2: Calculate Maximum Height
At maximum height, the vertical velocity becomes zero. Using the kinematic equation:
\[
v_y^2 = v_{0y}^2 - 2gh
\]
where:
- \( v_y = 0 \, \text{m/s} \) (final vertical velocity at maximum height),
- \( v_{0y} = 24.5 \, \text{m/s} \),
- \( g = 9.8 \, \text{m/s}^2 \).

Rearrange to solve for \( h \) (maximum height):
\[
0 = (24.5)^2 - 2 \times 9.8 \times h
\]
\[
h = \frac{(24.5)^2}{2 \times 9.8} = \frac{600.25}{19.6} \approx 30.64 \, \text{m}
\]

### Step 3: Calculate Time of Flight
The time to reach maximum height is given by:
\[
t_{\text{up}} = \frac{v_{0y}}{g} = \frac{24.5}{9.8} \approx 2.5 \, \text{s}
\]

Since the time going up equals the time coming down, the total time of flight is:
\[
t_{\text{total}} = 2 \times t_{\text{up}} = 2 \times 2.5 = 5 \, \text{s}
\]

### Step 4: Calculate Horizontal Distance (Range)
The horizontal distance \( R \) is given by:
\[
R = v_{0x} \times t_{\text{total}} = 42.44 \times 5 = 212.2 \, \text{m}
\]

### Final Answer
- **Maximum Height**: \( 30.64 \, \text{m} \)
- **Horizontal Distance (Range)**: \( 212.2 \, \text{m} \)

Would you like further explanation on any part?

### Related Questions
1. How does changing the launch angle affect the range and height?
2. What if the arrow was shot with a different initial velocity but the same angle?
3. How long would it take for the arrow to reach its peak height?
4. How would air resistance affect the arrow’s height and range?
5. Can this method apply if the angle is 90 degrees?

#### Tip
In projectile motion, remember that horizontal and vertical motions are independent but share the same time of flight.

If an arrow is shot at 30 degrees with the horizontal and has a velocity of 49 m/s, how high will the arrow go and what horizontal distance will the arrow travel?

Explore the projectile motion of an arrow shot at a 30-degree angle with an initial velocity of 49 m/s. Learn how to calculate the maximum height of 30.64 m and a horizontal range of 212.2 m using trigonometry and kinematics. Suitable for high school physics students.

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