The man lands 12.1 m far from the base of the cliff, so he will survive
Explanation:
The motion of the person is a projectile motion, 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, to find the time of flight of the person, with the following suvat equation:
[tex]s=ut+\frac{1}{2}gt^2[/tex]
where :
s = 50 m is the vertical displacement of the man (the height of the cliff)
t is the time of flight
u = 0 is the initial vertical velocity
[tex]g=9.8 m/s^2[/tex] is the acceleration of gravity
Solving for t, we find
[tex]t=\sqrt{\frac{2s}{g}}=\sqrt{\frac{2(50)}{9.8}}=3.19 s[/tex]
Now we know that the man moves horizontally, with a constant velocity of
[tex]v_x = 3.8 m/s[/tex]
Therefore, the horizontal distance covered by the man during this time is
[tex]d=v_x t = (3.8)(3.19)=12.1 m[/tex]
So the man lands 12.1 m far from the base of the cliff: therefore, on the lake, so he will survive.
Learn more about projectile motion:
brainly.com/question/8751410
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