2. A 0.8 kg tetherball hangs on the end of a cord. It is hit by a child and rises 2.1 m above the ground. a. What is the maximum
1 answer:
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Answer:
about 19.6° and 73.2°
Explanation:
The equation for ballistic motion in Cartesian coordinates for some launch angle α can be written ...
y = -4.9(x/s·sec(α))² +x·tan(α)
where s is the launch speed in meters per second.
We want y=2.44 for x=50, so this resolves to a quadratic equation in tan(α):
-13.6111·tan(α)² +50·tan(α) -16.0511 = 0
This has solutions ...
tan(α) = 0.355408 or 3.31806
The corresponding angles are ...
α = 19.5656° or 73.2282°
The elevation angle must lie between 19.6° and 73.2° for the ball to score a goal.
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I find it convenient to use a graphing calculator to find solutions for problems of this sort. In the attachment, we have used x as the angle in degrees, and written the function so that x-intercepts are the solutions.
Answer: acceleration:
velocity:
Explanation:
The complete question is written bellow:
<em>A cat is moving at 18 m/s when it accelerates at </em><em> for 2 seconds. What is his new velocity? </em>
<em />
In this situation the following equation will be useful:
Where:
is the cat’s final velocity (new velocity)
is the cat’s initial velocity
is the cat's acceleration
is the time
Solving the equation:
This is the cat's new velocity