The stone lands 133 m far from the bottom of the cliff
Explanation:
The motion of the stone in this problem is a projectile motion, so it follows a parabolic path which consists of two independent motions:
- A uniform motion (constant velocity) along the horizontal direction
- An accelerated motion with constant acceleration (acceleration of gravity) in the vertical direction
We start by considering the vertical motion: since it is a free fall motion, we can use the suvat equation
[tex]s=ut+\frac{1}{2}at^2[/tex]
where
s = 85 m is the vertical displacement, the height of the cliff (we chose downward as positive direction)
u = 0 is the initial vertical velocity of the stone
t is the time the stone takes to reach the ground
[tex]a=g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for t, we find the time:
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(85)}{9.8}}=4.16 s[/tex]
Now we can consider the horizontal motion, which is a uniform motion at constant speed. The horizontal distance travelled is given by
:
[tex]d=v_x t[/tex]
where we know:
[tex]v_x = 32 m/s[/tex] is the horizontal velocity, which is constant
t = 4.16 s is the time of flight
Solving for d, we find:
[tex]d=(32)(4.16)=133 m[/tex]
So, the stone lands 133 m far from the base of the cliff.
Learn more about projectile motion here:
brainly.com/question/8751410
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