contestada

An object of unknown mass is initially at rest and dropped from a height h. It reaches the ground with a velocity v1 v 1 . The same object is then raised again to the same height h but this time is thrown downward with velocity v1 v 1 . It now reaches the ground with a new velocity v2 v 2 . How is v2 v 2 related to v1 v 1 ?

Respuesta :

xero099

Answer:

[tex]\frac{v_{2}}{v_{1}} = \sqrt{2}[/tex]

Explanation:

Each object is modelled as follows:

First object

[tex]v_{1} = \sqrt{2\cdot g\cdot h}[/tex]

Second object

[tex]v_{2} = \sqrt{v_{1}^{2}+2\cdot g \cdot h}[/tex]

[tex]v_{2} = \sqrt{4\cdot g \cdot h}[/tex]

The relation between [tex]v_{2}[/tex] and [tex]v_{1}[/tex] is:

[tex]\frac{v_{2}}{v_{1}} = \sqrt{2}[/tex]

onyebuchinnaji

The relationship between the two velocities [tex]v_1[/tex] and [tex]v_2[/tex] is [tex]v_2 = v_1 \sqrt{2}[/tex]

The given parameters;

  • initial velocity of the object, u = 0
  • height of the object = h
  • final velocity of the object, = [tex]v_1[/tex]
  • initial velocity for the second motion, = [tex]v_1[/tex]
  • final velocity of the object in the second motion, = [tex]v_2[/tex]

The relationship between the initial velocity and final velocity is given as follows;

For the first motion;

[tex]v_1 ^2 = v_0^2 + 2gh\\\\v_1^2 = 0 + 2gh\\\\v_1^2 = 2gh[/tex]

For second motion;

[tex]v_2^2 = v_1^2 + 2gh\\\\[/tex]

Relationship between the two velocities;

[tex]v_2^2 = v_1^2 + 2gh\\\\recall , for \ first \ motion, \ v_1 ^2 = 2gh\\\\v_2^2 = v_1^2 + v_1^2\\\\v_2^2 = 2v_1^2 \\\\v_2 = \sqrt{2v_1^2} \\\\v_2 = v_1\sqrt{2}[/tex]

Thus, the relationship between the two velocities [tex]v_1[/tex] and [tex]v_2[/tex] is [tex]v_2 = v_1 \sqrt{2}[/tex]

Learn more here: https://brainly.com/question/14620239

Otras preguntas