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A person absentmindedly walks off the edge of a tall cliff. They will fall 50 m into either
deadly rocks or a safe lake below the cliff. The lake is 12 meters away from the edge of
the cliff
If the person walks off the cliff at a constant velocity of 3.8 m/s.
will they survive? (yes or no) How far did they go?

Respuesta :

The man can survive, and he lands 12.1 m from the base of the cliff (into the lake)

Explanation:

The motion of the person is equivalent to the motion of a projectile, which consists of two separate 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 analyzing the vertical motion, to find the time it take for the person to reach the ground level. We can use the following suvat equation:

[tex]s=ut+\frac{1}{2}gt^2[/tex]

where

s = 50 m is the vertical distance covered (the height of the cliff)

u = 0 is the initial vertical velocity

[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity

t is the time of flight

Solving for t,

[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(50)}{9.8}}=3.19 s[/tex]

The person moved horizontally at a constant speed of

[tex]v_x = 3.8 m/s[/tex]

So, the horizontal distance covered by the man during his flight will be

[tex]d_x = v_x t = (3.8)(3.19)=12.1 m[/tex]

So, the man will land on the lake (which starts from 12 m), so he can survive.

Learn more about projectile motion:

brainly.com/question/8751410

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