Question

A lunar lander is making its descent to moon base i. the lander descends slowly under the retro-thrust of its descent engine. the engine is cut off when the lander is h = 5.5 m above the surface and has a downward speed of 0.8 m/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 1.6 m/s2.

Answer 1

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

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Further explanation

Acceleration is rate of change of velocity.

$\boxed {a = \frac{v - u}{t} }$

$\boxed {d = \frac{v + u}{2}~t }$

$\boxed {v^2 = u^2 + 2ad}$

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 = 5.5 m

initial downward speed of lander = u = 0.8 m/s

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

Asked:

final speed of the lander = v = ?

Solution:

$v^2 = u^2 + 2gh$

$v^2 = 0.8^2 + 2(1.6)(5.5)$

$v^2 = 18.24$

$v = \sqrt{18.24}$

$v \approx 4.3 \texttt{ m/s}$

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

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Answer details

Grade: High School

Subject: Physics

Chapter: Kinematics

Answer 2

V = final velocity
u = initial velocity
a = acceleration
s = displacement
Take downward direction as positive.
v^2 = u^2 + 2as
v^2 = 0.8^2 + 2(1.6)(5.5)
v^2 = 18.24
v = 4.27 m/s