An object that is dropped from a height H falls with a constant acceleration of g. The final
speed, v, just before it reaches the ground can be expressed as v = sqrt(2gh). By how much does the height need to change double the final velocity?

Question

05-09-2022
Physics

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An object that is dropped from a height H falls with a constant acceleration of g. The final
speed, v, just before it reaches the ground can be expressed as v = sqrt(2gh). By how much does the height need to change double the final velocity?

Sagot :

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pstnonsonjoku pstnonsonjoku · Answer 1 · 2022-09-14T11:11:03-04:00

The height has to change by a factor of 1.

What is the height?

We can obtain the height from which the object falls using the equations of kinematics under gravity. We can see that the object was dropped from a height thus u = 0 m/s.

We can the write;

v= √2gh

h1 = v^2/2g

When the final velocity is doubled, we have;

2v= √2gh

h2 =4v^2/2g

h2 = 2v^2/g

The height changes by;

2v^2/g - v^2/2g = v^2/g

Thus the height has to change by a factor of 1.

Learn more about height of an object:https://brainly.com/question/14451550

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An object that is dropped from a height H falls with a constant acceleration of g. The final speed, v, just before it reaches the ground can be expressed as v = sqrt(2gh). By how much does the height need to change double the final velocity?

Sagot :

What is the height?

Other Questions

An object that is dropped from a height H falls with a constant acceleration of g. The final
speed, v, just before it reaches the ground can be expressed as v = sqrt(2gh). By how much does the height need to change double the final velocity?