A rocket is dropping to the moons surface at a constant rate of 15 feet per second. After 2 minutes of descent the rocket is at a height of 11,400 feet above the moons surface. At what elevation in feet was the rocket when it started to drop to the moons surface?

Question

Mathematics

Skylar Long

4 May, 17:29

A rocket is dropping to the moons surface at a constant rate of 15 feet per second. After 2 minutes of descent the rocket is at a height of 11,400 feet above the moons surface. At what elevation in feet was the rocket when it started to drop to the moons surface?

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Answers (1)

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Lawson Fleming · Answer 1

13,200 feet

Step-by-step explanation:

So, the rocket starts descending at a given rate and time, so we first need to find the length it passed during that descent.

Length equals to speed multiplied with time:

s = v • t

since speed is given in feet per second, we need to convert minutes to seconds; if one minute has 60 seconds, then two minutes have 120 seconds.

s = 15 feet/s • 120 s

s = 1800 feet

So, the rocket descended 1800 feet, but it still has 11,400 feet to the Moon's surface. That means that the rocket was on 11,400 + 1,800 = 13,200 feet elevation when it started to drop.