Answered

IDNStudy.com, kung saan ang iyong mga tanong ay may malinaw na sagot. Alamin ang mga detalyadong sagot mula sa mga bihasang miyembro ng aming komunidad na sumasaklaw sa iba't ibang paksa para sa lahat ng iyong pangangailangan.

A marble rolls off horizontally from the edge of a tabletop 1.50 m above the floor. It strikes the floor 2.0 m from the base of the table. How long does it take the marble to reach the floor? What is its initial speed?

Sagot :

Glendzmoralesidn

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

#LetsStudy

Ang iyong presensya ay mahalaga sa amin. Magpatuloy sa pagtatanong at pagbahagi ng iyong nalalaman. Sama-sama tayong magtutulungan upang makamit ang mas mataas na antas ng karunungan. Gawin mong pangunahing mapagkukunan ang IDNStudy.com para sa maasahang mga sagot. Nandito kami para sa iyo.