Problem 25 - College Physics

Question

. A water balloon slingshot launches its projectiles essentially from ground level at a speed of 25.0 \(\mathrm{m} / \mathrm{s}\) . (a) At what angle should the slingshot be aimed to achieve its maximum range? (b) If shot at the angle you calculated in part (a), how .far will a water balloon travel horizontally? (c) For how long will the balloon be in the air? (You can ignore air resistance.)

Step-by-Step Solution

Answer

(a) The angle is 45°. (b) The range is approximately 63.78 meters. (c) The time of flight is about 3.61 seconds.

Step 1: Determine the Angle for Maximum Range

In projectile motion, the angle for maximum range is achieved at \(45^\circ\). This is because, at this angle, the horizontal and vertical components of velocity are equal, providing the maximum range.

Step 2: Calculate the Horizontal Range

Using the formula for the range \( R \) of a projectile, \( R = \frac{v_0^2 \sin(2\theta)}{g} \), where \( v_0 = 25 \text{ m/s} \), \( \theta = 45^\circ \), and \( g = 9.8 \text{ m/s}^2 \).Calculate as follows:\[ R = \frac{(25)^2 \sin(90^\circ)}{9.8} = \frac{625}{9.8} \approx 63.78 \text{ meters} \]

Step 3: Calculate the Time of Flight

The time of flight \( T \) can be found using \( T = \frac{2v_0 \sin(\theta)}{g} \).Substitute the values:\[ T = \frac{2 \times 25 \times \sin(45^\circ)}{9.8} \]Given \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \), we have:\[ T = \frac{50 \times \frac{\sqrt{2}}{2}}{9.8} = \frac{50 \times 0.707}{9.8} \approx 3.61 \text{ seconds} \]

Key Concepts

Maximum Range

In projectile motion, achieving the maximum range is all about finding the perfect launch angle. For any projectile launched on Earth, this ideal angle is universally accepted to be \(45^\circ\). At this angle, the projectile's velocity is evenly split between its vertical and horizontal components. This symmetry allows it to travel the farthest possible distance before returning to the ground. The physics behind this phenomenon is beautifully simple. At \(45^\circ\), the projectile spends just enough time in the air while covering an extensive horizontal distance. This happens because both the upward path and downward path are equally long, maximizing the covered distance.

It’s helpful to visualize this by thinking of a triangle. When the sides that represent vertical and horizontal components of velocity are equal (as in the \(45^\circ\) angle triangle), the resultant trajectory efficiently stretches over the optimum range of land.

Horizontal Range

Once the ideal launch angle of \(45^\circ\) is determined, the next step is to calculate how far the projectile will travel horizontally. This distance is known as the **horizontal range**. The formula to find this is:

• \( R = \frac{v_0^2 \sin(2\theta)}{g} \)

Plug in the values given in the problem, where \(v_0 = 25 \text{ m/s}\), \(\theta = 45^\circ\), and \(g = 9.8 \text{ m/s}^2\), to solve the calculation:

• \( R = \frac{(25)^2 \times \sin(90^\circ)}{9.8} \)
• \( R = \frac{625}{9.8} \approx 63.78 \text{ meters} \)

The horizontal range of 63.78 meters tells us how far the projectile will land once it's launched at a \(45^\circ\) angle without any air resistance hindering its flight.

Time of Flight

The **time of flight** is the total duration the projectile spends in the air. This is an important aspect of projectile motion as it influences the range and the way a projectile behaves after being launched. You can calculate time of flight using the formula:

• \( T = \frac{2v_0 \sin(\theta)}{g} \)

Substitute in the values from our exercise session, where \(v_0 = 25 \text{ m/s}\), \(\theta = 45^\circ\), and \(g = 9.8 \text{ m/s}^2\):

• \( T = \frac{2 \times 25 \times \sin(45^\circ)}{9.8} \)
• With \(\sin(45^\circ) = \frac{\sqrt{2}}{2}\), compute as follows:
• \( T = \frac{50 \times 0.707}{9.8} \approx 3.61 \text{ seconds} \)

Thus, the \(3.61\) seconds tells us the water balloon will be airborne for 3.61 seconds. The formula hinges on the vertical component, demonstrating how long gravity acts on the projectile before returning it to ground level.

Angle of Projection

The **angle of projection** is key in defining the whole trajectory of a projectile. This exercise specifically deals with an angle that grants the projectile its maximum possible horizontal range, that is \(45^\circ\). By adjusting the angle of projection, the path and distance covered can vastly change.

It is crucial to understand that a steeper angle (greater than \(45^\circ\)) would lead to higher altitudes but shorter distances. Meanwhile, a shallower angle (less than \(45^\circ\)) makes a low trajectory that may not travel far because it’s not in the air long enough. Here’s why:

• At angles greater than \(45^\circ\), the vertical component of the velocity dominates, leading to more time in the air but falling short on the horizontal extent.
• Conversely, at angles less than \(45^\circ\), it’s the horizontal component that's emphasized, but gravity brings the projectile down too quickly, limiting range.

In this light, \(45^\circ\) presents a balanced approach where both components of the initial velocity effectively split their contributions, providing an ideal mix for maximum range.