SOLUTION: The window of a house is 6m above ground. A rocket is set off from the ground at an initial velocity of 20m/s.
For how many seconds will the rocket be at or above the window?
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-> SOLUTION: The window of a house is 6m above ground. A rocket is set off from the ground at an initial velocity of 20m/s.
For how many seconds will the rocket be at or above the window?
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Question 1208736: The window of a house is 6m above ground. A rocket is set off from the ground at an initial velocity of 20m/s.
For how many seconds will the rocket be at or above the window?
Use the equation
S(T)= -4.9T^2 + VoT + So to model the position S for a given time T, where Vo is the initial Velocity. Answer by math_tutor2020(3817) (Show Source):
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V0 = initial velocity = 20 meters per second
s0 = initial height = 0 meters because it was launched on the ground
Using a graphing tool like GeoGebra or Desmos (among many other options), you should get the following graph
y = -4.9x^2+20x is the parabola
y = 6 is the horizontal line
The parabola and horizontal line intersect at points A and B.
A = (0.326,6)
B = (3.756,6)
The decimal values are approximate. Round them however your teacher instructs.
Another way to find these decimal solutions is to use the quadratic formula to solve -4.9x^2 + 20x - 6 = 0.
Therefore, the rocket's height is at or above 6 meters on the interval
This is the portion of the curve between A and B.
This duration is roughly 3.43 seconds because 3.756-0.326 = 3.43
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