Answer:
The time to take the marble to reach the floor is 0.553 s and its initial speed is 5.422 m/s.
Explanation:
Free falling body is the motion of a falling object under the influence of the Earth's gravity. Its motion is independent of its weight. The constant acceleration of a free falling body is called acceleration due to gravity, denoted by [tex]g[/tex] which is approximate to 9.8 m/[tex]s^{2}[/tex].
For the formula to be used in the problem, we use:
[tex]V_{1} ^{2}=V_{0} ^{2} +2gy[/tex] equation 1
[tex]y=V_{0} t+\frac{1}{2} gt^{2}[/tex] equation 2
where
[tex]V_{1}[/tex] is the final velocity, unit is in m/s
[tex]V_{0}[/tex] is the initial velocity, unit is in m/s
[tex]t[/tex] is the time, unit is in seconds (s)
[tex]y[/tex] is the vertical distance, unit is in meters (m)
[tex]g[/tex] is the acceleration due to gravity, [tex]9.8 m/s^{2}[/tex]
For the given information
[tex]y=1.50m[/tex] height of the table
[tex]x=2.0m[/tex] distance of the marble from the base of the table (when the
marble strikes the floor)
Solving the problem
1. To solve for the initial velocity, use equation 1 then substitute the given information.
[tex]V_{1} ^{2}=V_{0} ^{2} +2gy[/tex]
[tex]0=V_{0} ^{2} +2(-9.8m/s^{2} )(1.5m)[/tex] [tex]g[/tex] is negative since the acceleration is
downward
[tex]V_{0} =\sqrt{(2(9.8)(1.5)}[/tex]
[tex]V_{0} =5.422m/s[/tex]
2. To solve for the time, use equation 2 then substitute the value of
[tex]V_{0} =5.422 m/s[/tex] and the given, we get:
[tex]y=V_{0} t+\frac{1}{2} gt^{2}[/tex]
[tex]1.50m=(5.422m/s)t+\frac{1}{2} (9.8)t^{2}[/tex]
Simplifying and arranging the equation
[tex]4.9t^{2} +5.422t-1.50=0[/tex]
Solving for [tex]t[/tex] using quadratic equation, [tex]x=\frac{-b+\sqrt{b^{2}-4ac } }{2a}[/tex], we get:
[tex]t=\frac{-5.422+\sqrt{5.422^{2+4(4.9)(1.50)} } }{2(4.9)}[/tex]
[tex]t=0.553sec[/tex]
For more information related to free falling body, just click on the following links:
* Recommendations about a free falling body
https://brainly.ph/question/2162209
* Additional example
https://brainly.ph/question/2170448
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