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#27 question

A penny is dropped by a parachuter and takes 5s to hit the ground. What is the altitude of the parachuter, assuming
g= 9.81 m/s² and ignoring air resistance?

A. 5m
B. 49m
C. 123m
D. 245m​

27 QuestionA Penny Is Dropped By A Parachuter And Takes 5s To Hit The Ground What Is The Altitude Of The Parachuter Assumingg 981 Ms And Ignoring Air Resistance class=

Sagot :

MDKPfromIDN

So, the initial altitude of the parachuter is approximately (C). 123 m.

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 following equations apply:

The equation for calculating the height (h)

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

The equation for calculating the time (s)

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

The equation for calculating the velocity (v)

[tex] \boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}} [/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)
  • v = velocity (m/s)

Problem Solving

We know that :

  • t = interval of the time = 5 s
  • g = acceleration of the gravity = 9.81 m/s²

What was asked :

  • h = height or displacement at vertical line = ... m

Step by Step :

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

[tex] \sf{h = \frac{1}{2} \cdot 9.81 \cdot 5^2} [/tex]

[tex] \sf{h = \frac{245.25}{2}} [/tex]

[tex] \boxed{\sf{h = 122.625 \: m \approx 123 \: m}} [/tex]

Conclusion

So, the initial altitude of the parachuter is approximately 123 m (C.)

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