MHB When Will the Object Be 15 Meters Above the Ground?

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An object propelled vertically upward with an initial velocity of 20 meters per second follows the equation s = -4.9t^2 + 20t. To determine when it reaches 15 meters above the ground, the equation simplifies to 4.9t^2 - 20t + 15 = 0, yielding two solutions: approximately 0.99 seconds while ascending and 3.09 seconds while descending. The discussion emphasizes the importance of accurately accounting for the gravitational acceleration, which is approximately 9.81 m/s². The calculations demonstrate the object's trajectory and the significance of quadratic equations in motion analysis. Understanding these principles is crucial for accurately predicting the object's behavior in vertical motion scenarios.
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
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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