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#26 question



An astronaut jumps to a height of 5.00m above the surface of the moon. If the acceleration due to gravity on the moon is 1.6 m/s², how long does it take for the astronaut to fall to the surface of the moon?


A. 1.01s
B. 2.48s
C. 3.09s
D. 6.17s ​

Sagot :

MDKPfromIDN

So, the time that taken for the astronaut to fall to the surface of the moon is 2.5 s.

Introduction

Hi ! In this question, I will help you. In this question, you will learn about the fall time of the free fall motion. Free fall is a downward vertical motion without being preceded by an initial velocity. When moving in free fall, the time required can be calculated by the following equation:

[tex] \sf{h = \frac{1}{2} \cdot g \cdot t^2} [/tex]

[tex] \sf{\frac{2 \cdot h}{g} = t^2} [/tex]

[tex] \boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}} [/tex]

With the following condition :

  • t = interval of the time (s)
  • h = height or any other displacement at vertical line (m)
  • g = acceleration of the gravity (m/s²)

Problem Solving

We know that :

  • h = height = 5.00 m
  • g = acceleration of the gravity = 1.6 m/s²

What was asked :

  • t = interval of the time = ... s

Step by step :

[tex] \sf{t = \sqrt{\frac{2 \cdot h}{g}}} [/tex]

[tex] \sf{t = \sqrt{\frac{2 \cdot 5}{1.6}}} [/tex]

[tex] \sf{t = \sqrt{6.25}} [/tex]

[tex] \boxed{\sf{t = 2.5 \: s}} [/tex]

Conclusion

So, the time that taken for the astronaut to fall to the surface of the moon is 2.5 s.

See More

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