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

Lucia Travis

22 March, 00:19

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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Ralph · 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.