Question 1208731
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All three statements are correct.
After the substitution, you can use the quadratic formula to find t.


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Example question: when will the object be 20 meters above the ground?


Solution:
s = -4.9t^2 + 20t
20 = -4.9t^2 + 20t
4.9t^2 - 20t + 20 = 0


Plug a = 4.9, b = -20, c = 20 into the quadratic formula
{{{t = (-b+-sqrt(b^2-4ac))/(2a)}}}


{{{t = (-(-20)+-sqrt((-20)^2-4(4.9)(20)))/(2(4.9))}}}


{{{t = (20+-sqrt(8))/(9.8)}}}


{{{t = (20+sqrt(8))/(9.8)}}} or {{{t = (20-sqrt(8))/(9.8)}}}


{{{t = 2.329431}}} or {{{t = 1.752201}}}
Both decimal results are approximate.


The object reaches the 20 meter mark at around 1.752 seconds, rises higher a little bit, then falls back down to the 20 meter mark at around 2.329 seconds


Another approach is to graph the parabola y = -4.9x^2 + 20x and the horizontal line y = 20
{{{
drawing(500,500,-2,5,-5,25,
graph(500,500,-2,5,-5,25,-100,-4.9x^2 + 20x),
line(-10,20,10,20),

circle(1.752201,20,0.02),circle(1.752201,20,0.03),circle(1.752201,20,0.04),circle(1.752201,20,0.05),circle(2.329431,20,0.02),circle(2.329431,20,0.03),circle(2.329431,20,0.04),circle(2.329431,20,0.05),

locate(1.752201-0.08,20-0.5,"A"),locate(2.329431-0.05,20-0.5,"B"),
locate(1.5,5.8,"A = (1.752201,20)"),
locate(1.5,4.8-0.5,"B = (2.329431,20)")
)
}}}
The two intersect at points A and B shown in the graph above. 
The x coordinates of these intersections are the approximate solutions found earlier.


<a href="https://www.geogebra.org/calculator">GeoGebra</a> and <a href="https://www.desmos.com/calculator">Desmos</a> are two graphing options, among many others. 
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