Problem 66 - University Physics with Modern Physics

Question

Henrietta is jogging on the sidewalk at 3.05 m/s on the way to her physics class. Bruce realizes that she forgot her bag of bagels, so he runs to the window, which is 38.0 m above the street level and directly above the sidewalk, to throw the bag to her. He throws it horizontally 9.00 s after she has passed below the window, and she catches it on the run. Ignore air resistance. (a) With what initial speed must Bruce throw the bagels so that Henrietta can catch the bag just before it hits the ground? (b) Where is Henrietta when she catches the bagels?

Step-by-Step Solution

Answer

Bruce throws the bagels at 2.88 m/s, and Henrietta is 35.94 meters past the window when she catches them.

Step 1: Determine Time for Bagels to Fall

First, we need to calculate the time it takes for the bagels to fall from Bruce's window to the ground. Since the bagels are thrown horizontally, we consider only vertical motion to determine this time. Using the equation for free fall: \[ s = \frac{1}{2}gt^2 \] where \( s = 38.0 \) m and \( g = 9.81 \) m/s². Re-arranging the equation, we get: \[ t = \sqrt{\frac{2s}{g}} \] Substitute the values: \[ t = \sqrt{\frac{2 \times 38.0}{9.81}} \] Solving this gives \( t \approx 2.78 \) seconds.

Step 2: Calculate Initial Speed for Horizontal Throw

To find the speed at which Bruce must throw the bagels horizontally, we need to consider Henrietta's position. The bagels must reach Henrietta in the same time it takes them to fall, which we've calculated as 2.78 seconds. Since Henrietta is running at 3.05 m/s, and she grabs the bagels on the run, Bruce's horizontal throw speed \( v_x \) must cover the same horizontal distance.Since Henrietta was running for 9 + 2.78 seconds: \( 11.78 \) seconds Henrietta covers: \[ d = 3.05 \times 11.78 \] Thus, Bruce's throw should also cover 3.05 m/s \( \times \) 2.78 s = \( 8.47 \) m in horizontal distance. His throw speed should be \[ v_x = \frac{8.47}{2.78} \approx 2.88 \text{ m/s} \].

Step 3: Find Henrietta's Position at Catch

Since Henrietta catches the bagels on the run, we calculate how far she has moved in total since she passed below the window. This is calculated by multiplying her speed with the total time from when Bruce threw the bagels until she catches them: \[ \text{Total time} = 9 + 2.78 = 11.78 \text{ seconds} \] \[ \text{Distance} = 3.05 \times 11.78 \] Solving gives \( 35.94 \) meters from the point directly below the window.

Key Concepts

Horizontal Motion

In projectile motion, horizontal movement is crucial to understand. This type of motion occurs without the influence of gravity impacting its speed horizontally.
Consider the bagel thrown horizontally from a height. It moves forward, and its path traces out a parabolic shape. The horizontal velocity remains constant because gravity does not affect horizontal motion. Thus, any change in this velocity results purely from initial throw force or external factors.
For our scenario, Bruce needs to ensure the bagel reaches Henrietta within the time frame calculated. Therefore, Bruce's horizontal throw is crucial, with his goal being for the bagel to travel exactly over the calculated horizontal distance throughout the fall.

Free Fall

Free fall relates to the bagel's vertical descent due to gravity's pull. In free fall, an object's only influence is gravity, causing it to accelerate downwards.
For our exercise, Bruce's location is 38 m above ground. Hence, he must calculate the time taken for the bagel to fall directly downwards.
The formula used: \[ s = \frac{1}{2}gt^2 \] Here, \(s\) is the fall distance (38 m), and \(g\) is acceleration due to gravity (9.81 m/s²). Solving, we find that the bagel's fall duration is about 2.78 seconds.
Bruce needs this insight to synchronize the throw with Henrietta's speed in horizontal motion.

Initial Speed Calculation

Determining the proper initial speed is essential for Bruce's successful throw. This speed ensures that the bagel reaches Henrietta just above ground level.
The factor here is Henrietta's velocity of 3.05 m/s. Bruce must calculate how swiftly the bag must travel in the horizontal direction during its fall.
With the fall lasting approximately 2.78 seconds, the horizontal distance is also a measurement to consider. Thus, Bruce figures out he needs to throw the bag at around 2.88 m/s. This speed ensures Henrietta can catch it "on the run," completing the task successfully.

Kinematics Equations

Understanding kinematics equations helps dissect the motion of the bagel in both directions. These equations apply classical mechanics to analyze the different components of motion separately.
For vertical motion, the equation assists in finding the fall time using: \[ t = \sqrt{\frac{2s}{g}} \] By solving, it shows how long the fall lasts by considering gravity's effect.
In terms of horizontal motion, the importance is using the equation: \[ d = vt \] This shows the relationship between distance, speed, and time for movements.
These equations collectively explain how Bruce calculates the throw and kinematic paths for successful delivery to Henrietta.