To solve this problem, let's break down the information provided:

1. Range ($R$): 200 meters
2. Time of flight ($T$): 7 seconds
3. Initial and constant acceleration due to gravity ($g$): 9.8 m/s²

Step-by-Step Solution

1. Calculate the Horizontal Component of the Initial Velocity ($v_{0x}$)

Since the projectile reaches the ground level at a range of 200 meters after 7 seconds, we can calculate the horizontal component of the initial velocity using the range formula: $R = v_{0x} \cdot T$ Solving for $v_{0x}$: $v_{0x} = \frac{R}{T} = \frac{200 \, \text{m}}{7 \, \text{s}} \approx 28.57 \, \text{m/s}$

2. Calculate the Vertical Component of the Initial Velocity ($v_{0y}$)

The time to reach the peak of the trajectory is half the total time of flight, so: $t_{\text{peak}} = \frac{T}{2} = \frac{7 \, \text{s}}{2} = 3.5 \, \text{s}$ Using the vertical motion equation at the peak (where final vertical velocity $v_y = 0$): $v_y = v_{0y} - g \cdot t_{\text{peak}}$ $0 = v_{0y} - (9.8 \, \text{m/s}^2)(3.5 \, \text{s})$ Solving for $v_{0y}$: $v_{0y} = (9.8 \, \text{m/s}^2)(3.5 \, \text{s}) = 34.3 \, \text{m/s}$

3. Calculate the Resultant Initial Velocity ($v_0$)

The initial velocity $v_0$ is the vector sum of the horizontal and vertical components: $v_0 = \sqrt{v_{0x}^2 + v_{0y}^2}$ $v_0 = \sqrt{(28.57)^2 + (34.3)^2} \approx 44.64 \, \text{m/s}$

4. Calculate the Angle of Launch ($\theta$)

The angle of launch $\theta$ can be found using the inverse tangent of the ratio of $v_{0y}$ to $v_{0x}$: $\theta = \tan^{-1} \left( \frac{v_{0y}}{v_{0x}} \right)$ $\theta = \tan^{-1} \left( \frac{34.3}{28.57} \right) \approx 50.19^\circ$

5. Calculate the Maximum Height ($H$)

Using the formula for the maximum height in projectile motion: $H = \frac{v_{0y}^2}{2g}$ $H = \frac{(34.3)^2}{2 \times 9.8} \approx 60.04 \, \text{m}$

Summary of Results

• Horizontal Component of Initial Velocity ($v_{0x}$): $\approx 28.57 \, \text{m/s}$
• Vertical Component of Initial Velocity ($v_{0y}$): $\approx 34.3 \, \text{m/s}$
• Resultant Initial Velocity ($v_0$): $\approx 44.64 \, \text{m/s}$
• Angle of Launch ($\theta$): $\approx 50.19^\circ$
• Maximum Height ($H$): $\approx 60.04 \, \text{m}$

Would you like further details on any of these steps, or do you have any additional questions?

Here are five related questions to deepen understanding:

1. How would changing the launch angle affect the range of the projectile?
2. What would be the impact on maximum height if the initial speed were increased?
3. How does the time of flight change with different angles of launch for the same initial speed?
4. What would happen to the horizontal and vertical components if the projectile were launched at a steeper angle?
5. How is the range affected if the projectile is launched from a height instead of ground level?

Tip: When solving projectile motion problems, always start by breaking down the initial velocity into its horizontal and vertical components, as this simplifies calculations across different parts of the motion.