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.