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;
Therefore, for the correct option, we have;
Learn more about velocity of an object under gravity here:
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