When the person kicks a rock horizontally off a cliff we have:
a) The height of the cliff is:
[tex]y_{f}=706.32\: m[/tex]
b) The final velocity of the rock before it hits the ground is:
[tex]v_{yf}=-117.72\: m/s[/tex]
c) The range of the rock is:
[tex]x=420\: m[/tex]
a)
We have here a semi-parabolic motion. We can use the following equation to find the height of the cliff.
[tex]y=y_{i}+v_{iy}t-0.5gt^{2}[/tex]
The initial velocity if the y-direction is zero is this motion. So we have:
[tex]y=y_{i}-0.5gt^{2}[/tex]
The initial height (y(i) = 0) and the final height will be negative.
[tex]-y_{f}=-0.5gt^{2}[/tex]
[tex]y_{f}=0.5*9.81*12^{2}[/tex]
[tex]y_{f}=706.32\: m[/tex]
Therefore, the height the cliff is 706.32 m.
b)
The final vertical velocity will be:
[tex]v_{yf}=v_{yi}-gt[/tex]
We know that the initial velocity if the y-direction is zero so we will have:
[tex]v_{yf}=-gt[/tex]
[tex]v_{yf}=-9.81*12[/tex]
[tex]v_{yf}=-117.72\: m/s[/tex]
Therefore, the final velocity is -117.72 m/s.
c)
Finally, the range can be found using this equation:
[tex]v_{x}=\frac{x}{t}[/tex]
We know the velocity in the x-direction is constant. Therefore, the range (x) will be:
[tex]x=v_{x}t[/tex]
[tex]x=35*12[/tex]
[tex]x=420\: m[/tex]
Therefore, the range will be 420 m.
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You can learn more about parabolic motion here:
https://brainly.com/question/24240176
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