Question

A quarterback can throw a receiver a high, lazy lob pass or a low, quick bullet pass. These passes are indicated by curves 1 and 2 , respectively, in Figure 4.43. (a) The lob pass is thrown with an initial speed of \(21.5 \mathrm{~m} / \mathrm{s}\), and its time of flight is \(3.97 \mathrm{~s}\). What is its launch angle? (b) The bullet pass is thrown with a launch angle of \(25.0^{\circ}\). What is the initial speed of this pass? (c) What is the time of flight of the bullet pass?

Short answer

(a) \(\theta \approx 42.92^\circ\), (b) Solve for \(v_i\), (c) Solve for \(t\).

Step-by-step solution

Understanding the Problem

This problem involves projectile motion. We need to find the launch angle of a lob pass, initial speed of a bullet pass, and time of flight for the bullet pass using physics equations for projectile motion.

Find Launch Angle (Part a)

For the lob pass, we know initial speed \(v_i = 21.5 \ \text{m/s}\) and time of flight \(t = 3.97 \ \text{s}\). We use the equation \( t = \frac{2v_i \sin(\theta)}{g} \) where \(g = 9.81 \ \text{m/s}^2\) to find \(\theta\).Rearranging for \(\theta\):\[\theta = \arcsin\left(\frac{gt}{2v_i}\right)\]Plugging in the given values:\[\theta = \arcsin\left(\frac{9.81 \times 3.97}{2 \times 21.5}\right) \approx 42.92^\circ\]

Find Initial Speed (Part b)

For the bullet pass, we know the launch angle \(\theta = 25.0^{\circ}\). The time of flight equation is derived from horizontal motion:\[x_f = v_i \cos(\theta) \cdot t\]Solving for initial speed gives:\[v_i = \frac{x_f}{\cos(\theta) \cdot t}\]We also use the vertical motion equation:\[y_f = v_i \sin(\theta) \cdot t - \frac{1}{2}gt^2 = 0\]Using these two equations, solve for \(v_i\) (usually requires numerical or graphical methods since \(t\) is not given).

Find Time of Flight (Part c)

Using the equation derived from vertical motion:\[y_f = v_i \sin(\theta) \cdot t - \frac{1}{2}gt^2 = 0\]The time of flight \(t\) for a projectile launched from and returning to the same height is:\[t = \frac{2v_i \sin(\theta)}{g}\]Substituting \(v_i\) from part b (once calculated), solve for \(t\).

Key concepts

Launch Angle Determination

To determine the launch angle in projectile motion, we have to focus on the vertical component of the motion. The time of flight equation used in this context is \[ t = \frac{2v_i \sin(\theta)}{g} \], where:

• \(t\) is the time of flight,
• \(v_i\) is the initial speed, and
• \(g\) is the acceleration due to gravity (approximately \(9.81 \ \text{m/s}^2\)).

We rearrange this formula to solve for the launch angle \(\theta\):\[ \theta = \arcsin\left(\frac{gt}{2v_i}\right) \].Plugging in the values provided in the original problem and using inverse sine, we can calculate that the lob pass is thrown with a launch angle of approximately \(42.92^\circ\).
This concept highlights how initial velocity and gravity interact to influence the angle needed for achieving a specific time of flight.

Initial Speed Calculation

Calculating the initial speed of a projectile, such as a bullet pass in the exercise, involves understanding both horizontal and vertical components of the motion. Given the launch angle \(\theta\), the vertical motion can be analyzed using \[ y_f = v_i \sin(\theta) \cdot t - \frac{1}{2} gt^2 = 0 \].
The horizontal motion, often not involving the influence of gravity in its immediate calculation, uses:\[ x_f = v_i \cos(\theta) \cdot t \].
Here’s the process:

• Identify horizontal and vertical displacement equations.
• Note that for a typical projectile, final vertical position \(y_f\) is often zero (assuming same height at launch and landing).
• Use the known factors to solve the equations (often involving simultaneous equations when \(t\) isn't directly provided).

While the exact initial speed of the bullet pass requires numerical or graphical methods since \(t\) isn't given explicitly, these equations provide the foundation needed for solving more complex problems.

Time of Flight Analysis

The time of flight is a fundamental component of projectile motion analysis. It signifies the duration for which a projectile remains airborne. The basic equation to find the time of flight for a projectile returning to its launch height is:\[ t = \frac{2v_i \sin(\theta)}{g} \],where all variables mean the same as before.
To determine the time of flight, one must:

• Consider both the initial speed \(v_i\) and the angle of launch \(\theta\), as these will influence how long the projectile remains aloft.
• Remember that when a projectile lands at the same height from which it was launched, the formula simplifies as shown.
• For the bullet pass, once the initial speed is known from prior computations, substitute back to get the exact value of \(t\).

Understanding these interactions makes it possible to predict motion scenarios, crucial for both direct calculations and higher-level physics applications.