Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Time in the air of the marble using the equation y=1/2gt^2, re-arranged to make t the subject - t = √(2y/g). Therefore… t = √(0.94x2/9.81) = 0.4377…
where g=9.81.
Using the equation x=Ut (the horizontal displacement) 2.35=u x 0.4377..
re-arrange to make u = x/t. Therefore.. u=2.35/0.4377 = 5.368m/s
=5.37 = 3 significant figures
where g=9.81.
Using the equation x=Ut (the horizontal displacement) 2.35=u x 0.4377..
re-arrange to make u = x/t. Therefore.. u=2.35/0.4377 = 5.368m/s
=5.37 = 3 significant figures
The initial velocity of the marble as it leaves the launching device is [tex]\boxed{5.35\text{ m/s}}[/tex] .
Further explanation:
When an object is thrown horizontally from a certain height, the object moves both in x and y direction due to the gravitational force.
Given:
The height of the table is [tex]94.0\text{ cm}[/tex].
The object strikes the ground at [tex]2.35\text{ m}[/tex].
Concept:
Consider the east direction as the positive x axis and vertical upward direction as the positive y axis.
During flight, the object is subjected to a downward acceleration [tex]g[/tex].
In horizontal direction the external force on the object is zero. Therefore, the acceleration in horizontal direction will be zero.
The motion of object in both x and y direction is:
In vertically upward direction.
The initial velocity of the marble is zero and an acceleration, in vertically downward direction, is acting on it.
[tex]{u_{\text{y}}}=0[/tex]
[tex]{a_{\text{y}}}=-g[/tex]
The displacement of the marble in the vertical direction is:
[tex]\begin{aligned}{s_{\text{y}}}&=- h \\&=- 0.94\,{\text{m}} \\ \end{aligned}[/tex]
The second equation of motion is:
[tex]{s_{\text{y}}}={u_{\text{y}}}t+\dfrac{1}{2}{a_{\text{y}}}{t^2}[/tex]
Substitute [tex]0\,{\text{m/s}}[/tex] for [tex]{u_y}[/tex], [tex]-h[/tex] for [tex]{s_{\text{y}}}[/tex] and [tex]-g[/tex] for [tex]{a_y}[/tex] in the above expression.
[tex]h=\frac{1}{2}g{t^2}[/tex]
Rearrange the above expression for [tex]t[/tex].
[tex]\boxed{t=\sqrt {\dfrac{{2h}}{g}}}[/tex]
Substitute [tex]0.94\text{ m}[/tex] for [tex]h[/tex] and [tex]9.81\,{\text{m/}}{{\text{s}}^{\text{2}}}[/tex] for [tex]g[/tex] in the above expression.
[tex]\begin{aligned}t&=\sqrt {\frac{{2\left( {0.94\,{\text{m}}} \right)}}{{\left( {9.81\,{\text{m/}}{{\text{s}}^{\text{2}}}} \right)}}} \\&=0.437\,{\text{s}} \\ \end{aligned}[/tex]
In horizontal direction.
[tex]{a_x}=0[/tex]
The displacement of marble in the horizontal direction is:
[tex]{s_{\text{x}}}=2.35\,{\text{m}}[/tex]
The second equation of motion is:
[tex]\boxed{{s_{\text{x}}}={u_{\text{x}}}t+\frac{1}{2}{a_{\text{x}}}{t^2}}[/tex]
Substitute the values in above equation.
[tex]2.35={u_{\text{x}}}\left( {0.437\,{\text{s}}} \right)+0[/tex]
Simplify the above expression for [tex]{u_{\text{x}}}[/tex].
[tex]\begin{aligned}{u_{\text{x}}}&=\frac{{2.35}}{{0.437}} \\&=5.35\,{\text{m/s}} \\ \end{aligned}[/tex]
In x direction the acceleration is zero. Therefore, velocity of marble will remain same in horizontal or x direction throughout the motion.
Thus, the initial velocity of the marble as it leaves the launching device is [tex]\boxed{5.35\text{ m/s}}[/tex] .
Learn More:
1. Charge on the capacitor https://brainly.com/question/8892837
2. Wavelength of the radiation https://brainly.com/question/9077368
3. Conservation of energy brainly.com/question/3943029
Answer Details:
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords:
Marble, 94.0 cm, 0.94 m, 2.35 m, initial velocity, launching device, projectile, 5.37 m/s, 5.37 meter per second, time of strike, 0.437 s, table, second equation of motion.
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Thanks for using our platform. We aim to provide accurate and up-to-date answers to all your queries. Come back soon. Westonci.ca is committed to providing accurate answers. Come back soon for more trustworthy information.
