To solve this problem, we need to break the motion into two components: vertical and horizontal.

Given:

• Initial velocity $v_0 = 8 \, \text{m/s}$
• Launch angle $\theta = 50^\circ$
• Acceleration due to gravity $g = 9.8 \, \text{m/s}^2$

Step 1: Vertical Component of Motion

The vertical component of the initial velocity is: $v_{0y} = v_0 \sin(\theta) = 8 \sin(50^\circ)$ Now calculate $v_{0y}$.

Step 2: Maximum Height

At the maximum height, the vertical velocity becomes 0. Using the kinematic equation: $v_y^2 = v_{0y}^2 - 2g h$ Where $v_y = 0$ at the maximum height: $0 = v_{0y}^2 - 2g h$ Solving for $h$, the maximum height: $h = \frac{v_{0y}^2}{2g}$

Step 3: Time of Flight

To find the total time the ball is in the air, use the formula for the time to reach the maximum height, and then double it (since the time to rise and fall are the same): $t_{\text{up}} = \frac{v_{0y}}{g}$ Total time of flight: $t_{\text{total}} = 2 t_{\text{up}} = 2 \times \frac{v_{0y}}{g}$

Step 4: Horizontal Distance (Range)

The horizontal component of the velocity is: $v_{0x} = v_0 \cos(\theta) = 8 \cos(50^\circ)$ The horizontal distance traveled is: $R = v_{0x} \times t_{\text{total}}$

Let me now calculate the maximum height $h$ and the range $R$.### Results:

• The maximum height the ball reaches is approximately 1.92 meters.
• The horizontal distance traveled before hitting the ground is approximately 6.43 meters.

Would you like more details, or do you have any questions?

Here are 5 related questions to deepen your understanding:

1. What would happen to the maximum height if the launch angle were increased to 60°?
2. How does the range change if the initial velocity increases to 10 m/s at the same angle?
3. What is the significance of the horizontal and vertical components in projectile motion?
4. How can air resistance affect the calculations and results of this problem?
5. How would the time of flight change if the launch angle were 30°?

Tip: The angle for maximum range in projectile motion, assuming no air resistance, is 45°.