Respuesta :
Let the height of the chin is H from the floor
so the time taken by the ball to reach the bottom on the surface of earth is given by kinematics
[tex]H = \frac{1}{2} gt^2[/tex]
now we will have time from above equation
[tex]t = \sqrt{\frac{2H}{g}}[/tex]
here on earth the gravity is to be g = 10 m/s^2
now we will have
[tex]T_{earth} = \sqrt{\frac{2H}{10}}[/tex]
now for the surface of other planet time is given by
[tex]T_{planet} = \sqrt{\frac{2H}{30}}[/tex]
now we can find the ratio of two time
[tex]T_{earth} : T_{planet} = \sqrt3[/tex]
it shows that on the other planet where gravity is more the ball will take less time to reach the surface and will acquire more velocity when it will hit the floor
The ball will take lesser time to reach the ground on the surface of the planet having value of acceleration due to gravity [tex]30\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}[/tex] by a factor of [tex]\fbox{\begin\\\sqrt 3\end{minispace}}[/tex].
Further Explanation:
Since the ball is dropped from a particular height on the planet more massive than the Earth, the ball will have initial velocity to be zero as it starts from rest.
The ball further falls through the distance under the action of acceleration due to gravity on the planet. The value of acceleration due to gravity is different for the different planets depending on their sizes.
Concept:
The expression for the ball falling a height H under the action of the acceleration due to gravity is:
[tex]H = {v_i}t + \dfrac{1}{2}g{t^2}[/tex] ……. (1)
Here, [tex]{v_i}[/tex] is the initial velocity of the ball, [tex]g[/tex] is the acceleration due to gravity on the surface of the planet and [tex]t[/tex] is the time taken by the ball to fall.
On the surface of Earth:
The value of acceleration due to gravity on the surface of Earth is approximately [tex]10\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}[/tex]
Substitute [tex]0[/tex] for [tex]{v_i}[/tex], [tex]10\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}[/tex] for [tex]g[/tex] and [tex]{t_{Earth}}[/tex] for [tex]t[/tex] in equation (1).
[tex]\begin{aligned} {t_{Earth}}& = \sqrt {\frac{{2H}}{g}}\\&= \sqrt {\frac{{2H}}{{10}}} \,{\text{s}} \\ \end{aligned}[/tex]
On the surface of other planet:
The value of acceleration due to gravity on the other planet is [tex]30\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}[/tex]
Substitute [tex]0[/tex] for [tex]{v_i}[/tex], [tex]30\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}[/tex] for [tex]g[/tex] and [tex]{t_{Planet}}[/tex] for [tex]t[/tex] in equation (1).
[tex]\begin{aligned}{t_{Planet}}&=\sqrt{\frac{{2H}}{g}}\\&=\sqrt{\frac{{2H}}{{30}}}\,{\text{s}}\\\end{aligned}[/tex]
Compare the time taken by the ball to cover the height H on two surfaces.
[tex]\begin{aligned}\frac{{{t_{Planet}}}}{{{t_{Earth}}}}&=\frac{{\sqrt{\frac{{2H}}{{30}}}}}{{\sqrt{\frac{{2H}}{{10}}}}}\\&=\frac{1}{{\sqrt3}}\\\end{aligned}[/tex]
Therefore, the above expression suggests that the ball takes [tex]\fbox{\begin\\{\sqrt 3}\end{minispace}}[/tex] times more time on the surface of Earth to fall through the height H as compared to that on the surface of the planet with the value of acceleration due to gravity equal to [tex]30\,{{\text{m}} \mathord{\left/ {\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}[/tex].
Learn More:
1. The translational kinetic energy of a gas molecule https://brainly.com/question/9078768
2. The expansion of a gas or Charles' Law https://brainly.com/question/9979757
3. Stress developed in a wire brainly.com/question/12985068
Answer Details:
Grade: College
Subject: Physics
Chapter: Kinematics
Keywords:
Ball, height, H, Earth, other planet, g=30m/s2, g=30, acceleration due to gravity, root3, (3)^1/2, size, planet, rest, initial speed, g=10m/s2.
