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Question

The first word spoken on the surface of the Moon after Apollo 15 landed on July 30, 1971, was “Bam!” this was James Irwin’s involuntary reaction to their rather bone-jarring touchdown. “We hit harder than any other flights!” says Irwin. “And I was startled, obviously, when I said, ‘Bam!’” The reason for the “firm” touchdown of Apollo 15 was that the rocket engine was shit off when the lander was still 1.31 m above the lunar surface and moving downward with a speed of 0.152 m/s. From that point on the lander descended in a lunar free fall, with an acceleration of 1.62 m/s2. As a result, the landing speed of Apollo 15 was by far the largest of any of the Apollo missions. To visualize the descent of Apollo 15, consider its position-time curve during the final stages of landing, shown below. 

a. How much time did it take for the lander to drop the final 1.31 m to the Moon”s surface? 

b. What was the impact speed of the lander when it touched down?

Answer 1

First, I hope that you noticed the hilarious typo in the problem when the "rocket engine was shut off".

Second, it isn't the "proper physics" way to solve part a, but you can just read the graph. t_shutoff=8.7s and t_touchdown=9.8s, so the difference is 1.1 seconds. If nothing else this can be used to check the answer to a.

a. Use the kinematic equation: y = .5at²+v₀t+y₀ [it is common to see this with x or d instead of y, but I like to switch it to y for vertical problems, also the 0.5 is usually written as 1/2, and some will write the terms in reverse order, but algebra's order of operations tells us it doesn't matter the order that we add them in]. We use this one because we want to solve for time and are given a height, initial velocity, and acceleration. We also know that when it gets to the surface the height will be 0 m.

0 m = .5 x (-1.62 m/s²) x t² + (-0.152 m/s) x t +1.31 m - since the acceleration and initial velocities are downward they are negative.

Then use the quadratic equation to solve and you get t = -1.36 s and 1.18 s, since 1) the impact happens in the future and not the past the negative time can be ignored and 2) the positive time is close to what we got from reading the graph, then the t=1.18 s must be the correct answer. We will use this answer in part b.

b) use the kinematic equation: v = at+v₀ [you will sometimes see f and i for final and initial in place of no subscript and 0 on the velocities]. We use this one because we want to know v and have an acceleration, time (solved from part a), and initial velocity. We could use others, but this is the shortest and simplest of the kinematic equations given what we know.

v = -1.62 m/s² x 1.18s + -0.152 m/s

v = -2.06 m/s , where the negative sign simple means that the velocity was in the downward direction.