Kinematics - How long does this basketball take to fall down

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Homework Help Overview

The problem involves kinematics, specifically the motion of a basketball tossed vertically upward with an initial speed of 4.6 m/s. The question focuses on determining the minimum time a player must wait before touching the ball after it reaches its maximum height and begins to fall.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss initial conditions such as initial height, velocity, and acceleration due to gravity. Questions are raised about the characteristics of the basketball at maximum height and the relationship between speed, acceleration, and time.

Discussion Status

The discussion is ongoing, with some participants providing guidance on the known variables and prompting further exploration of the problem's dynamics. There is no explicit consensus yet, as various interpretations and approaches are being considered.

Contextual Notes

Participants are working within the constraints of the problem as posed, including the initial conditions and the requirement that the player cannot touch the ball until it begins to fall. The discussion includes questioning the signs used in the acceleration due to gravity.

maryyy16
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I have no idea how to even begin this problem:

At the beginning of a basketball game, a referee tosses the ball straight up with the speed of 4.6 m/s. A player cannot touch the ball until after it reaches its maximum height and begins to fall down. What is the minimum time that a player must wait before touching the ball?

PLEASE HELP!
 
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First, take an inventory of what you know. Initial position? Well, we can just agree to call the initial height zero. Initial velocity? Yes: 4.6 m/s, with the upward direction being positive. Acceleration? Yes, it's the acceleration of gravity, a = -9.8 m/s2. (Do you know why there's a minus sign?)

Now, what about position, velocity, acceleration, or time characterizes the basketball when it's at its maximum height?
 
As the ball is thrown up it slows down at a rate of 'g' = 9.8m/s^2
At the top of the curve it has a speed of 0.
What is the relationship between speed, acclearation and time ?
 
u have the formula: x=0.5*a*(t^2)+v(0) where x is the maximum heigh and a is 9.8 and t is the time and v is the initial speed. just calculate it
 

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