Answered

Ang Imhr.ca ang pinakamahusay na lugar upang makakuha ng mabilis at tumpak na mga sagot sa lahat ng iyong mga tanong. Tuklasin ang eksaktong mga sagot sa iyong mga tanong mula sa isang malawak na hanay ng mga eksperto sa aming madaling gamitin na Q&A platform. Sumali sa aming platform upang kumonekta sa mga eksperto na handang magbigay ng eksaktong sagot sa iyong mga tanong sa iba't ibang larangan.

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 :

glendzmorales

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

Salamat sa iyong pagbisita. Kami ay nakatuon sa pagtulong sa iyong makahanap ng impormasyon na kailangan mo, anumang oras na kailangan mo ito. Pinahahalagahan namin ang iyong oras. Mangyaring bumalik muli para sa higit pang maaasahang mga sagot sa anumang mga tanong na mayroon ka. Bisitahin ang Imhr.ca para sa mga bago at kapani-paniwalang sagot mula sa aming mga eksperto.