Q40E - University Physics with Modern Physics

Question

A lunar lander is making its descent to Moon Base I (Fig. E2.40). The lander descends slowly under the retro-thrust of its descent engine. The engine is cut off when the lander is 5.0 m above the surface and has a downward speed of$\mathbf{0}\mathbf{.}\mathbf{8}\mathbf{ }\mathbf{m}\mathbf{/}\mathbf{s}$. 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$\mathbf{1}\mathbf{.}\mathbf{6}\mathbf{ }\mathbf{m}\mathbf{/}{\mathbf{s}}^{\mathbf{2}}$.

 

Step-by-Step Solution

Answer

The speed of the lander just before it touches the surface is$=4.08 \mathrm{m}/\mathrm{s}$.

Step 1: Given data

Distance from which lander experiences free fall due to gravity is$h=5.0 \mathrm{m}$

 

The downward speed of the lander is$u=0.8 \mathrm{m}/\mathrm{s}$  

 

The acceleration due to gravity at the moon is$a=1.6 \mathrm{m}/{\mathrm{s}}^2$  

Step 2: Solution

Newton’s third law of motion can be expressed as, 

$v^2=u^2+2ah$ 

 

Where $v$is the final velocity which is before touching the ground

 

Substituting these values in the above expression gives,

 

 $$\begin{gathered} V^2={\left(0.8 \mathrm{m}/\mathrm{s}\right)}^2\times \left(5.0 \mathrm{m}\right) \\ {V}^2={\left(4.08 \mathrm{m}/\mathrm{s}\right)}^2 \\ {V}^2=4.08 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

Thus, the final velocity of the lunar lander before touching the surface is$V^2=4.08 \mathrm{m}/\mathrm{s}$