The height of the cliff is 96.2 m
Explanation:
The motion of the rock along the horizontal direction is a uniform motion (constant velocity), since there are no forces acting along that direction. Therefore, the horizontal velocity is given by
[tex]v_x = \frac{d}{t}[/tex]
where, using the data of the first rock:
d = 66.5 m is the horizontal distance travelled by the rock
[tex]v_x = 15 m/s[/tex] is the horizontal velocity
Solving for t, we find the time it takes for the rock to reach the ground:
[tex]t=\frac{d}{v_x}=\frac{66.5}{15}=4.43 s[/tex]
The vertical motion instead is a free fall motion acted upon gravity, so we can use the suvat equation:
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s is the vertical displacement of the rock, which corresponds to the height of the cliff
u = 0 is the initial vertical velocity of the rock
t = 4.43 s is the time of flight
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity (we chose downward as positive direction)
Solving for s, we find the height of the cliff:
[tex]s=0+\frac{1}{2}(9.8)(4.43)^2=96.2 m[/tex]
Learn more about projectile motion:
brainly.com/question/8751410
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