The speed of the lander just before it touches the surface is about 4.3 m/s
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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 = 5.5 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:
[tex]v^2 = u^2 + 2gh[/tex]
[tex]v^2 = 0.8^2 + 2(1.6)(5.5)[/tex]
[tex]v^2 = 18.24[/tex]
[tex]v = \sqrt{18.24}[/tex]
[tex]v \approx 4.3 \texttt{ m/s}[/tex]
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The speed of the lander just before it touches the surfacet is about 4.3 m/s
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Grade: High School
Subject: Physics
Chapter: Kinematics
