Q40E - University Physics with Modern Physics

Question

A lunar lander is makingits descent to Moon Base I (Fig. E2.40). The lander descendsslowly under the retro-thrust of its descent engine. The engine iscut off when the lander is 5.0 m above the surface and has a downwardspeed of 0.8m/s . With the engine off, the lander is in freefall. What is the speed of the lander just before it touches the surface?The acceleration due to gravity on the moon is $1.6m/s^2$ .

Step-by-Step Solution

Answer

The speed of the lander just before it touchesthe surface is $v=08m/s$$v=4.08m/s$.

Step 1: Given data

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

 The downward speed of the lander is  $u=0.8m/s$

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

Step 2: Solution

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

$v^2=u^2+2ah$

Where  is the final velocity which is before touching the ground

Substituting these values in the above expression gives,

$$\begin{gathered} v^2={\left(0.8m/s\right)}^2+2\left(1.6m/s^2\right)\times \left(5.0m\right) \\ {v}^2={\left(4.08m/s\right)}^2 \\ v=4.08m/s \end{gathered}$$

 Thus, the final velocity of the lunar lander before touching the surface is  $v=4.08m/s$