Respuesta :
The time taken by the hammer to fall on the surface of the moon is [tex]\boxed{1.57\,{\text{s}}}[/tex].
Further Explanation:
Given:
The acceleration due to gravity on the surface of the moon is [tex]1.62\,{{\text{m}} \mathord{\left/{\vphantom {{\text{m}} {{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}[/tex].
The height from which the hammer falls towards the surface is [tex]2\,{\text{m}}[/tex].
Concept:
The hammer falls freely without action of any force or any friction of the medium. It will fall freely under the acceleration due to gravity on the surface of the moon.
The motion of the hammer freely under the gravity is governed by the second equation of motion. The second equation of motion can be written as:
[tex]\boxed{S=ut+\frac{1}{2}a{t^2}}[/tex]
Here, [tex]S[/tex] is the distance through which the hammer falls, [tex]u[/tex] is the initial speed of the hammer and [tex]a[/tex] is the acceleration due to gravity on the surface of moon.
Since the hammer is just dropped, the initial velocity of the hammer will be zero.
Substitute the values in the above expression.
[tex]\begin{aligned}2&=\left( {0 \times t}\right)+\left( {\frac{1}{2}\times 1.62 \times {t^2}}\right) \\t&=\sqrt {\frac{{2 \times 2}}{{1.62}}} \,{\text{s}}\\&= 1.{\text{57}}\,{\text{s}}\\\end{aligned}[/tex]
Thus, the time taken by the hammer to fall on the surface of the moon is [tex]\boxed{1.57\,{\text{s}}}[/tex].
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Answer Details:
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords: Astronaut, drops a hammer, surface of moon, acceleration due to gravity, equation of motion, time taken by hammer, to fall, force, friction, freely under gravity.
If an astronaut drops a hammer from 2.0 meters above the surface of the moon, it will take the 1.57 seconds hammer to fall to the moon's surface.
Given the following data:
- Displacement = 2.0 meters
- Acceleration due to gravity on the moon = 1.62 [tex]m/s^2[/tex]
- Initial velocity = 0 m/s
To find how long (time) it will take for the hammer to fall to the moon's surface, we would use the second equation of motion.
Mathematically, the second equation of motion is given by the formula;
[tex]S = ut + \frac{1}{2} at^2[/tex]
Where:
- S is the displacement.
- u is the initial velocity.
- a is the acceleration.
- t is the time measured in seconds.
Substituting the values into the formula, we have;
[tex]2 = 0t + \frac{1}{2} (1.62)t^2\\\\2 = 0.81t^2\\\\t^2 = \frac{2}{0.81} \\\\t^2 = 2.47\\\\t = \sqrt{2.47}[/tex]
Time, t = 1.57 seconds.
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