Three identical rocks are launched with identical speeds from the top of a platform of height Oh. The rocks are launched in the directions indicated above.
Which of the following correctly relates the magnitude vy of the vertical component of the velocity of each rock immediately before it hits the ground?

Respuesta :

Answer:

All rocks hit the ground with the same speed.

Explanation:

The magnitude of the vertical velocity under gravity changes by approximately 9.81 m/s each second

The option that correctly relates the magnitude, [tex]v_{y}[/tex], of the vertical velocity is the option;

[tex](v_{y1} = v_{y1}) \geq v_{y3}[/tex]

Reason:

The possible question options are; [tex](v_{y1} = v_{y1}) \geq v_{y3}[/tex], [tex]v_{y1}> v_{y3} > v_{y2}[/tex], [tex]v_{y2}> v_{y3} > v_{y1}[/tex], [tex]v_{y1}= v_{y2} = v_{y3}[/tex]

Known parameters;

The speed of the rocks = u

The height of the platform from which the rocks are thrown = h₀

The given direction in which the balls are thrown are;

Rock 1; Direction  = 45° above the horizontal

Rock 2; Direction = 45° below the horizontal

Rock 3; Direction = 0° or in the horizontal direction

The vertical component of the speed of Rock 1= -u·sin(45°)

Vertical component of the speed of Rock 2 = u·sin(45°)

Vertical component of the speed of Rock 3 = u·sin(0°) = 0

The velocity of the rock is given by

v² = u² - 2·g·h₀

[tex]v = \pm \sqrt{u^2 - 2 \cdot g \cdot h_0}[/tex]

Where g is taken as negative, we have;

[tex]v = \pm \sqrt{u^2 + 2 \cdot g \cdot h_0}[/tex]

Therefore, the vertical velocity the rocks, Rock 1 and Rock 2, thrown with magnitude of velocity, [tex]|v_y| = |u \cdot sin(45^{\circ})|[/tex] is the same during the vertical motion of the ball at the same height, but change only in sign

[tex]At \ a \ given \ height, \ h_0, \ v_{y1} = v_{y2}[/tex]

[tex]Therefore, \ before \ the \ rocks \ hit \ the \ ground, \ v_{y1} = v_{y2}[/tex]

The initial vertical velocity of Rock 3, [tex]v_{y3}[/tex] = 0, therefore;

  • [tex]v_{y3} = \pm \sqrt{0 + 2 \cdot g \cdot h_0} \leq \pm \sqrt{u^2 + 2 \cdot g\cdot h_0 } = (v_{y1} = v_{y1})[/tex]

Therefore, for the correct option, we have;

  • [tex](v_{y1} = v_{y1}) \geq v_{y3}[/tex]

Learn more about velocity of an object under gravity here:

https://brainly.com/question/14608286

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