Respuesta :
If there is a rock that is free falling from rest where acceleration due to gravity is 4.0 m/s^2, and it falls 32 meters, the velocity can be solved as follows:
velocity = +-sqrt(gravity x y displacement + initial velocity^2)
velocity = +-sqrt(4.0m/s^2 x 32 meters + 0^2)
velocity = +-sqrt(128m/s)
velocity = 11.3 m/s
The speed of the rock after it falls 32 meters is 11.3m/s
The speed of the rock after it has fallen [tex]32\text{ m}[/tex] near the surface of the planet is [tex]\boxed{16\text{ m/s}}[/tex].
Explanation:
Given:
The acceleration due to gravity on the surface of the planet is [tex]4.0\text{ m/s}^2[/tex].
The distance through which the rock falls is [tex]32\text{ m}[/tex].
Concept:
As the rock falls near the surface of he planet, it falls freely under the action of the acceleration due to gravity i.e. the gravitational pull of the planet acting on the rock.
The acceleration due to gravity is different on the surface of the different planets. It depends on the mass and radius of the particular planet.
The final speed of the rock after falling through a particular distance is given by the third equation of motion.
[tex]\boxed{v_f^2=v_i^2+2gS}[/tex]
Here, [tex]v_f[/tex] is the final velocity of rock, [tex]v_i[/tex] is the initial velocity, [tex]g[/tex] is the acceleration due to gravity on planet and [tex]S[/tex] is the distance covered by the rock.
As the rock falls from the rest, the initial velocity of the rock will be zero.
Substitute the values in above expression.
[tex]\begin{aligned}v_f^2&=(0)^2+2(4)(32)\\v_f&=\sqrt{256}\\&=16\text{ m/s}\end{aligned}[/tex]
Thus, The speed of the rock after it has fallen [tex]32\text{ m}[/tex] near the surface of the planet is [tex]\boxed{16\text{ m/s}}[/tex].
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Laws of motion
Keywords:
acceleration, gravity, rock, planet, surface, equation of motion, initial, final, distance, velocity, 32 m, 16 m/s.
