Tell whether the statement is always, sometimes, or never true.
1 answer:
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Answer:
The function that describes the height of the ball in time is .
Step-by-step explanation:
Let suppose that ball experiments a free fall motion, which means that the ball is accelerated because of gravity and gravitational acceleration can be considered constant since height reached by the object is too small in comparison with the radius of the Earth. Therefore, we can assume that ball is accelerated uniformly.
Hence, the kinematic formula for the height of the ball (), in feet, is described below:
(1)
Where:
- Initial height with respect to the ground, in feet.
- Initial velocity, in feet per second.
- Time, in seconds.
- Gravitational acceleration, in feet per square second.
If we know that ,
and
, then the function that describes the height of the ball in time is:
(2)
Comment
By similar triangles it can be shown that AD^2 = AB*AC
If you want the proof, Google tangents and secants of a circle.
Find
So we want
AB
Givens
AD = 16
BC = 9
AB = ??
CA = CB + AB
CA = 9 + AB
Formula
AB * (AB + BC) = AD^2
Sub and Solve
AB*(AB + 9) = 16^2
AB*(AB + 9) = 256 Remove the brackets.
AB^2 + 9AB = 256 Subtract 256 from both sides.
AB^2 + 9AB - 256 = 0
You can only do this either with a graph or the quadratic formula. I'll get the graph for you. You can made these yourself at Desmos.
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)
a = 1
b = 9
c = -256
Answer
When you substitute these into the quadratic formula, you get
x1 = 12.12 and
x2 = -21.12
x2 is meaningless. The solution is
x = 12.12
Comment
But that's not your question. Your question is what is this rounded to the nearest 1/10th? That's a fancy way of saying round to the first decimal place. Since the hundredth place (or second place) is 2, 12.12 rounds to 12.1
The answer is
x = 12.1 <<<<< answer.
By similar triangles it can be shown that AD^2 = AB*AC
If you want the proof, Google tangents and secants of a circle.
Find
So we want
AB
Givens
AD = 16
BC = 9
AB = ??
CA = CB + AB
CA = 9 + AB
Formula
AB * (AB + BC) = AD^2
Sub and Solve
AB*(AB + 9) = 16^2
AB*(AB + 9) = 256 Remove the brackets.
AB^2 + 9AB = 256 Subtract 256 from both sides.
AB^2 + 9AB - 256 = 0
You can only do this either with a graph or the quadratic formula. I'll get the graph for you. You can made these yourself at Desmos.
x = [-b +/- sqrt(b^2 - 4ac)] / (2a)
a = 1
b = 9
c = -256
Answer
When you substitute these into the quadratic formula, you get
x1 = 12.12 and
x2 = -21.12
x2 is meaningless. The solution is
x = 12.12
Comment
But that's not your question. Your question is what is this rounded to the nearest 1/10th? That's a fancy way of saying round to the first decimal place. Since the hundredth place (or second place) is 2, 12.12 rounds to 12.1
The answer is
x = 12.1 <<<<< answer.