A lunar lander is making its descent to the moon Base I. THe lander descends slowly under the reto-thrust of its descent engine. The engine is cut off when the lander is 5.0m above the surface and has a downward speed of 1.5m/s. With the engine off, the lander is in free fall. What is the speed of the lander just before it touches the surface? The acceleration due to gravity on the moon is 1.6m/s^2

Respuesta :

Answer: -4m/s

Explanation:

This problem is a good example of Free Fall, where the main condition is that the initial velocity must be zero [tex]V_{o}=0[/tex], and the main equations for this situation as follows:

[tex]y=y_{o}+V_{o}t-\frac{1}{2}gt^{2}[/tex]  (1)

[tex]V=V_{o}-gt[/tex]   (2)

Where:

[tex]y=0[/tex] is the final height of the lunar lander

[tex]y_{o}=5m[/tex]  is the initial height of the lunar lander (we are told the engine is cut off when the lander is 5.0m above the surface)

[tex]V_{o}=0[/tex] is the initial velocity of the lunar lander (we are told when the engine is off at 5m, the lander is in free fall)

[tex]t[/tex] is the time

[tex]g=1.6m/s^{2}[/tex] is the acceleration due to gravity on the moon

Having this clear, let's start finding [tex]t[/tex] from (1):

[tex]0=5m-\frac{1}{2}(1.6m/s^{2})t^{2}[/tex]  (3)

[tex]t=\sqrt{\frac{2(5m)}{1.6m/s^{2}}}[/tex]  (4)

[tex]t=2.5s[/tex]  (5)

Substituting (5) in (2):

[tex]V=0-gt=-(1.6m/s^{2})(2.5s)[/tex]   (6)

Finally:

[tex]V=-4m/s[/tex]  Note the velocity is negative because of the downward trajectory of the lunar lander due to the free fall.

The speed of the lander is about 4.3 m/s

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Further explanation

Acceleration is rate of change of velocity.

[tex]\boxed {a = \frac{v - u}{t} }[/tex]

[tex]\boxed {d = \frac{v + u}{2}~t }[/tex]

[tex]\boxed {v^2 = u^2 + 2ad}[/tex]

where:

a = acceleration ( m/s² )

v = final velocity ( m/s )

u = initial velocity ( m/s )

t = time taken ( s )

d = distance ( m )

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Let us now tackle the problem!

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Given:

position of the lander above the surface = h = 5.0 m

initial downward speed of lander = u = 1.5 m/s

acceleration due to gravity on the moon = g = 1.6 m/s²

Asked:

final speed of the lander = v = ?

Solution:

[tex]v^2 = u^2 + 2gh[/tex]

[tex]v^2 = 1.5^2 + 2(1.6)(5.0)[/tex]

[tex]v^2 = 18.25[/tex]

[tex]v = \sqrt{18.25}[/tex]

[tex]v \approx 4.3 \texttt{ m/s}[/tex]

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Conclusion :

The speed of the lander just before it touches the surfacet is about 4.3 m/s

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Learn more

  • Velocity of Runner : https://brainly.com/question/3813437
  • Kinetic Energy : https://brainly.com/question/692781
  • Acceleration : https://brainly.com/question/2283922
  • The Speed of Car : https://brainly.com/question/568302

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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

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