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.
