Question
What is the relationship between the final velocity of a dropped object and the height from which it is dropped, asked by a user named "wassaname_kal33ya"?
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Answered by
GPT-4o mini
The relationship between the final velocity of a dropped object and the height from which it is dropped can be described using the principles of physics, particularly free fall under the influence of gravity. When an object is dropped from a height, it accelerates downward due to gravity, and its final velocity can be calculated using the following kinematic equation:
\[ v^2 = u^2 + 2gh \]
where:
- \( v \) is the final velocity just before it hits the ground,
- \( u \) is the initial velocity (which is 0 if the object is dropped from rest),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \) near the Earth's surface),
- \( h \) is the height from which the object is dropped.
From this equation, if we assume that the object is dropped (not thrown), then \( u = 0 \). Therefore, the equation simplifies to:
\[ v^2 = 2gh \]
Rearranging this gives:
\[ v = \sqrt{2gh} \]
This shows that the final velocity \( v \) of the object just before impact is directly related to the square root of the height \( h \) from which it is dropped. Consequently, as the height increases, the final velocity just before impact also increases. This relationship assumes no air resistance or other forces acting on the object besides gravity.
\[ v^2 = u^2 + 2gh \]
where:
- \( v \) is the final velocity just before it hits the ground,
- \( u \) is the initial velocity (which is 0 if the object is dropped from rest),
- \( g \) is the acceleration due to gravity (approximately \( 9.81 \, m/s^2 \) near the Earth's surface),
- \( h \) is the height from which the object is dropped.
From this equation, if we assume that the object is dropped (not thrown), then \( u = 0 \). Therefore, the equation simplifies to:
\[ v^2 = 2gh \]
Rearranging this gives:
\[ v = \sqrt{2gh} \]
This shows that the final velocity \( v \) of the object just before impact is directly related to the square root of the height \( h \) from which it is dropped. Consequently, as the height increases, the final velocity just before impact also increases. This relationship assumes no air resistance or other forces acting on the object besides gravity.