Concept explainers
How long after the ball is released from the balcony, the friend has to wait, to start running so that she'll be able to catch the ball exactly 1.00 m above the floor of the court.

Answer to Problem 63QAP
The friend has to wait for 0.25 s after the ball is released and then start running so as to catch the ball at the height of 1.00 m above the floor.
Explanation of Solution
Given info:
The height of the balcony above the court
Initial velocity of the ball
Angle at which the ball is released
Distance of the initial position of the friend from the balcony
Friend's initial velocity
Friend's acceleration
Height at which the ball is caught
Formula used:
The equations of motion for vertical and horizontal motion of the ball can be used to find the time the friend needs to wait.
For vertical motion,
Here,
For horizontal motion of the ball,
Here,
The friend's motion can be analyzed using the equation,
Here,
Calculation:
Assume the origin A to be located at the point just below the balcony, with the x axis parallel to the ground and the positive y axis directed upwards. The height of the balcony from the ground is OA. The friend stands at B initially, and then catches the ball at the point C at a height CD from the ground. This is represented by the diagram shown below.
The ball is released with an initial velocity
The vertical motion of the ball is governed by the gravitational force. The acceleration of the ball in the vertical direction is equal to the acceleration of free fall.
Therefore,
The ball makes a vertical displacement from the initial position
Therefore,
Rewrite the equation (1) using the above expression, equation (4).
Substitute the values of the variables in the equation and calculate the value of time of flight t.
Rewrite the equation as a quadratic for t.
Solve for t.
Take the positive root alone.
The ball is in flight for 1.71 s. During this time, the ball travels a horizontal distance
Therefore,
Since point A is directly below the balcony, its x coordinate is
Therefore,
Use equation (5) and the values of the variables in the equation (2) and calculate the value of x.
The friend stands at the position
The horizontal displacement the friend needs to make is given by,
In equation (3), substitute the known values of the variables and calculate the time
Simplify the expression and solve for
The ball takes a time
Calculate the time
Conclusion:
Thus, the friend has to wait for 0.25 s after the ball is released and then start running so as to catch the ball at the height of 1.00 m above the floor.
Want to see more full solutions like this?
Chapter 3 Solutions
COLLEGE PHYSICS,VOLUME 1
- Hi Expert in Physics, I have uploaded pictures with respect to some physics equations. Could please name all Greek alphabet and their English name?arrow_forward81 SSM Figure 29-84 shows a cross section of an infinite conducting sheet carrying a current per unit x-length of 2; the current emerges perpendicularly out of the page. (a) Use the Biot-Savart law and symmetry to show that for all points B P P. BD P' Figure 29-84 Problem 81. x P above the sheet and all points P' below it, the magnetic field B is parallel to the sheet and directed as shown. (b) Use Ampere's law to prove that B = ½µλ at all points P and P'.arrow_forwardWhat All equations of Ountum physics?arrow_forward
- Please rewrite the rules of Quantum mechanics?arrow_forwardSuppose there are two transformers between your house and the high-voltage transmission line that distributes the power. In addition, assume your house is the only one using electric power. At a substation the primary of a step-down transformer (turns ratio = 1:23) receives the voltage from the high-voltage transmission line. Because of your usage, a current of 51.1 mA exists in the primary of the transformer. The secondary is connected to the primary of another step-down transformer (turns ratio = 1:36) somewhere near your house, perhaps up on a telephone pole. The secondary of this transformer delivers a 240-V emf to your house. How much power is your house using? Remember that the current and voltage given in this problem are rms values.arrow_forwardThe human eye is most sensitive to light having a frequency of about 5.5 × 1014 Hz, which is in the yellow-green region of the electromagnetic spectrum. How many wavelengths of this light can fit across a distance of 2.2 cm?arrow_forward
- A one-dimensional harmonic oscillator of mass m and angular frequency w is in a heat bath of temperature T. What is the root mean square of the displacement of the oscillator? (In the expressions below k is the Boltzmann constant.) Select one: ○ (KT/mw²)1/2 ○ (KT/mw²)-1/2 ○ kT/w O (KT/mw²) 1/2In(2)arrow_forwardTwo polarizers are placed on top of each other so that their transmission axes coincide. If unpolarized light falls on the system, the transmitted intensity is lo. What is the transmitted intensity if one of the polarizers is rotated by 30 degrees? Select one: ○ 10/4 ○ 0.866 lo ○ 310/4 01/2 10/2arrow_forwardBefore attempting this problem, review Conceptual Example 7. The intensity of the light that reaches the photocell in the drawing is 160 W/m², when 0 = 18°. What would be the intensity reaching the photocell if the analyzer were removed from the setup, everything else remaining the same? Light Photocell Polarizer Insert Analyzerarrow_forward
- The lifetime of a muon in its rest frame is 2.2 microseconds. What is the lifetime of the muon measured in the laboratory frame, where the muon's kinetic energy is 53 MeV? It is known that the rest energy of the muon is 106 MeV. Select one: O 4.4 microseconds O 6.6 microseconds O 3.3 microseconds O 1.1 microsecondsarrow_forwardThe Lagrangian of a particle performing harmonic oscil- lations is written in the form L = ax² - Bx² - yx, where a, and are constants. What is the angular frequency of oscillations? A) √2/a B) √(+2a)/B C) √√Ba D) B/αarrow_forwardThe mean temperature of the Earth is T=287 K. What would the new mean temperature T' be if the mean distance between the Earth and the Sun was increased by 2%? Select one: ○ 293 K O 281 K ○ 273 K 284 Karrow_forward
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781305952300Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- Physics for Scientists and EngineersPhysicsISBN:9781337553278Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningCollege PhysicsPhysicsISBN:9781938168000Author:Paul Peter Urone, Roger HinrichsPublisher:OpenStax CollegeCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage Learning





