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 0.8 m> 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>s 2 .

Answer 1

To calculate the speed of the lunar lander just before it touches the surface, the kinematic equation for linear motion under constant acceleration due to gravity is used, yielding an approximate speed of 3.92 m/s.

Step-by-step explanation:

To determine the speed of the lunar lander just before it touches the surface of the moon, we can use the kinematic equation for linear motion without constant acceleration due to gravity. We know the initial velocity (v0) is 0.8 m/s, the initial position (s0) is 5.0 m above the surface, the final position (s) is 0 m (when it touches the surface), and the acceleration (a) due to gravity on the moon is 1.6 m/s2.

The kinematic equation we will use is:

v2 = v02 + 2a(s - s0)

Plugging in the values, we have:

v2 = (0.8 m/s)2 + 2(1.6 m/s2)(0 m - 5.0 m)

Solving for v, we find:

v2 = 0.64 m2/s2 - 16 m2/s2

v2 = -15.36 m2/s2

Since v2 represents the square of the final velocity, we take the square root:

v = √(-15.36 m2/s2)

Since velocity cannot be negative when considering magnitude, we disregard the negative sign and have:

v = 3.92 m/s

Thus, the lander's speed just before touching the surface is approximately 3.92 m/s.