an object dropped off the top of a tall buildingfalls vertically with constant a
ID: 3092123 • Letter: A
Question
an object dropped off the top of a tall buildingfalls vertically with constant accerlation. if s is distance of theobject above the ground (in feet) t seconds after its relasethen s and t are related by an equation of the form s=a+bt^2 where a and b are constant suppose the object is 180feet above the ground 1 second after its release and 132 feet abovethe ground 2 seconds after its release (A) find the constant a and b (b) how high is the building (c)how long does the object fall an object dropped off the top of a tall buildingfalls vertically with constant accerlation. if s is distance of theobject above the ground (in feet) t seconds after its relasethen s and t are related by an equation of the form s=a+bt^2 where a and b are constant suppose the object is 180feet above the ground 1 second after its release and 132 feet abovethe ground 2 seconds after its release (A) find the constant a and b (b) how high is the building (c)how long does the object fallExplanation / Answer
The position (s) after 1 second (t=1) is 180, so the equationwould be 180 = a + b(1)^2, or simply 180 = a + b
Doing the same for the second part, your equation is 132 = a +b(2)^2, or 132 = a + 4b
Now you have a systems of equations, and you can solve for thetwo variables. If you need help with this process, justrespond to this =].
For part (b), you can figure out how high the building is bysubstituting time=0, because this is how high it will be withouthaving dropped it yet. So plug in 0 for t, and solve for sgiven the values of a and b that you found earlier.
For part (c), you want to find the time at which the object isat the ground, or when s=0. Plug in 0 for s, and solve fort!
I hope this helps! If you need any clarification, justask!
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