When Will the Object Be 15 Meters Above the Ground?

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Discussion Overview

The discussion revolves around the motion of an object propelled vertically upward with an initial velocity of 20 meters per second, specifically focusing on determining when the object will be 15 meters above the ground. Participants explore the mathematical modeling of the object's height over time, including potential calculations for when it strikes the ground and whether it reaches a height of 100 meters.

Discussion Character

  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents the equation for the object's height as $s = -4.9t^2 + 20t$ and attempts to solve for when $s = 15$ meters, leading to the equation $15 = -4.9t^2 + 20t$.
  • Another participant points out an error in the first participant's equation setup, noting that the term involving $t$ was dropped.
  • Subsequent calculations lead to the quadratic equation $4.9t^2 - 20t + 15 = 0$, which is further simplified to $49t^2 - 200t + 150 = 0$.
  • Estimates for the time when the object is at 15 meters are provided, with two potential solutions: $t \approx 3.0914$ seconds and $t \approx 0.99024$ seconds, with a suggestion that the latter is likely the time when the object is ascending.
  • There is a discussion about the acceleration due to gravity, with a participant noting that while $g$ is approximately $9.81$ m/s², a value of $9.8$ m/s² is often used for simplicity.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the correct interpretation of the time values calculated, and there is disagreement regarding the setup of the initial equation. The discussion remains unresolved with multiple viewpoints on the calculations and assumptions involved.

Contextual Notes

There are limitations in the discussion regarding the assumptions made about the acceleration due to gravity and the simplifications in the quadratic equation. The exact conditions under which the object is analyzed are not fully clarified.

karush
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$\tiny{1.2.1}$
An object is propelled vertically upward with an initial velocity of 20 meters per second.
The distance s (in meters) of the object from the ground after t seconds is
$s=-4.9t^2+20t$
(a) When will the object be 15 meters above the ground?
$15=-4.9t^2+20 \implies -4.9t^2 =-5$
ok there is no term b so decided not to use quadratic formula
so far...:unsure:
$49t^2=50$

(b) When will it strike the ground?
(c) Will the object reach a height of 100 meters
 
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karush said:
$\tiny{1.2.1}$
An object is propelled vertically upward with an initial velocity of 20 meters per second.
The distance s (in meters) of the object from the ground after t seconds is
$s=-4.9t^2+20t$
(a) When will the object be 15 meters above the ground?
$15=-4.9t^2+20 \implies -4.9t^2 =-5$
ok there is no term b so decided not to use quadratic formula
You dropped the t on the 20t term in going from [math]s = -4.9t^2 + 20t[/math] to [math]15 = -4.9t^2 + 20t[/math].

-Dan
 
.
 
Last edited:
topsquark said:
You dropped the t on the 20t term in going from [math]s = -4.9t^2 + 20t[/math] to [math]15 = -4.9t^2 + 20t[/math].

-Dan

$15 = -4.9t^2 + 20t
\implies 4.9t^2-20t+15=0
\implies 49t^2-200t+150=0$
kinda hefty for a quadratic equation so went to W|A
$t\approx 3.0914s$ probably this since it is going up
$t\approx 0.99024s $

it was tempting to just round off the 4.9 but think this how fast things fall
 
karush said:
$15 = -4.9t^2 + 20t
\implies 4.9t^2-20t+15=0
\implies 49t^2-200t+150=0$
kinda hefty for a quadratic equation so went to W|A
$t\approx 3.0914s$ probably this since it is going up
$t\approx 0.99024s $

it was tempting to just round off the 4.9 but think this how fast things fall
Mostly a good job. On the way up it passes 15 m at t = 0.099024 s. g is the acceleration due to gravity so it's how fast it is changing how fast it is falling. (Just call it an acceleration.. it's easier!)

Technically g is about 9.81 m/s^2 but the number locally is slightly different everywhere so it changes a bit. 9.8 m/s^2 is good enough.

-Dan
 

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