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The Apollo Lunar Module was used to make the transition from the spacecraft to the Moon's surface and back. Consider a similar module for landing on the surface of Mars. Use conservation of mechanical energy to answer these questions.

(a) As the lander is descending, if the pilot decides to shut down the engine when the lander is at a height of 2.7 m, (this may not be a safe height to shut down the engine) and the velocity of the lander (relative to the surface of the planet) is 1.8 m/s what will be velocity of the lander at impact? Note: g on the surface of Mars is about 0.4 times that on the surface of the Earth.

(b) In the case of the lunar module an impact velocity of 3.0 m/s or less was essential for a safe landing. Assuming this to be the case for the Mars lander as well, at what maximum height could the pilot shut down the engines to ensure a safe landing. Assume the velocity v0 at the time the engine is shut down is 1.8 m/s.

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boffeemadrid

Answer:

a) 4.94263 m/s

b) 0.734 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

g = Acceleration due to gravity

[tex]g=0.4\times 9.81\\\Rightarrow g=3.924\ m/s^2[/tex]

[tex]v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 3.924\times 2.7+1.8^2}\\\Rightarrow v=4.94263\ m/s[/tex]

The impact velocity of the rocket will be 4.94263 m/s

[tex]v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{3^2-1.8^2}{2\times 3.924}\\\Rightarrow s=0.734\ m[/tex]

The maximum height the pilot could shut down the engines to ensure a safe landing at initial velocity 1.8 m/s and final velocity 3 m/s is 0.734 m

abidemiokin

The velocity of the lander at impact is 4.94m/s

The maximum height the pilot could shut down the engines to ensure a safe landing is 0.735m

a) In order to get the velocity of the lander at impact, we will use Newton's law of motion expressed as:

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

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

s is the height of lander

Given the following parameters

u = 1.8m/s

g = 0.4 × 9.8 = 3.92m/s²

s = 2.7m

[tex]v^2=1.8^2+2(3.92)(2.7)\\v^2=3.24+21.168\\v^2=24.408\\v=\sqrt{24.408}\\v= 4.94m/s[/tex]

Hence the velocity of the lander at impact is 4.94m/s

b) To get the maximum height the pilot could shut down the engines to ensure a safe landing, we will use the formula:

[tex]v^2=u^2+2as\\3.0^2=1.8^2+2(3.92)s\\9.0-3.24=7.84s\\5.76=7.84s\\s=\frac{5.76}{7.84}\\s=0.735m[/tex]

Hence the maximum height the pilot could shut down the engines to ensure a safe landing is 0.735m

Learn more here: https://brainly.com/question/3385759

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