Stone is thrown from the top of the building.

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Homework Help Overview

The problem involves a stone being thrown downward from a height of 60 meters with an initial velocity of 20 m/s. The objective is to determine the time it takes for the stone to reach the ground.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of the height as a distance in the kinematic equation and question how to properly set up the equation with respect to the direction of motion.

Discussion Status

Some participants have attempted to set up the kinematic equation but express uncertainty about the signs used for the initial velocity and height. There is a suggestion to consider the direction of the stone's motion and its implications for the signs in the equation.

Contextual Notes

Participants are navigating the conventions of positive and negative values in the context of downward motion, which may affect their setup of the problem. There is also a hint provided regarding the use of the quadratic equation for solving for time.

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Homework Statement


A stone is thrown from the top of a building with an initial velocity of 20 m/s downward. The top of the building is 60 m above the ground. How much time elapses between the instant of release and the instant of impact with the ground?

Vi = -20 m/s height = 60 m
t = ?


Homework Equations



d = Vi.t + 1/2 a t2

The Attempt at a Solution


Couldn't figure out. How to use height as a distance?
 
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It has the same dimension ! The equations are OK.
 
So 60 m = -20m/s . t + 1/2 a t2
60 m = -20m/s t - 4.9 m/s2 t2?
What should I do after this ...
 
patelneel1994 said:
So 60 m = -20m/s . t + 1/2 a t2
60 m = -20m/s t - 4.9 m/s2 t2?
You seem to be using the convention that up is positive, which is fine. But is the stone traveling 60 m downward or upward? What does that tell you about the sign of the 60 m in your equation?

What should I do after this ...
Solve for t. (Hint: quadratic equation.)
 

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