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Understanding vertical motion equations is key to solving projectile motion problems like the water balloon exercise. Vertical motion accounts for the effects of gravity, which pulls objects downward. In physics, the vertical displacement \(\Delta y\) can be described by the equation:

• \(\Delta y = v_{0y}t - \frac{1}{2}gt^2\)

This formula shows how the position changes over time \(t\), with \(v_{0y}\) as the initial vertical velocity, and \(g\) gravitational acceleration (\(9.8\, \text{m/s}^2\)).

The equation helps us predict where a projectile lands by plugging in the known time and solving for initial conditions. In the exercise, the first balloon's \(\Delta y\) was \(-4 \text{ m}\), indicating it ended lower than its start point. Solving for \(v_{0y}\) (7.8 m/s) required manipulating the vertical motion equation. This approach is crucial for understanding how objects freefall under gravity's influence.

Horizontal motion in projectile problems differs from vertical because it lacks gravitational influence. In the water balloon problem, we used the formula:

• \(\Delta x = v_{0x}t\)

Here, \(\Delta x\) is horizontal displacement, and \(v_{0x}\) is the horizontal component of the initial velocity. With time \(t\) known, we calculate how far the object travels horizontally.

In the exercise, the horizontal distance was 35 meters. Since the balloon hit after 2 seconds, \(v_{0x}\) was obtained as 17.5 m/s. This component remained constant since no forces acted horizontally. Understanding this helps students predict an object's path parallel to the ground. This method emphasizes how horizontal and vertical motions combine in projectile scenarios.

Calculating initial velocity is a fundamental step in projectile motion. Combining vertical and horizontal components forms the total initial velocity. For the balloons to follow the desired path, precise initial velocities are necessary. In this context, we first dissected the velocity into vertical (\(v_{0y}\)) and horizontal (\(v_{0x}\)) parts.

For part (a), solving each component allowed us to use trigonometry to find the launch angle. Using \(\tan(\theta) = \frac{v_{0y}}{v_{0x}}\), we obtained the angle \(\theta \approx 24.5^\circ\). This illustrates the balance of both components.

In part (b), calculating \(v_0\) required combining components: \(v_0 = \sqrt{(v_{0x})^2 + (v_{0y})^2}\). The resulting speed of 22.0 m/s highlights how initial conditions affect projectile motion. Having a solid grasp of this concept is essential for predicting trajectories accurately.