Question

#27 question

Answer 1

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)

$\boxed{\sf{\bold{h = \frac{1}{2} \cdot g \cdot t^2}}}$

The equation for calculating the time (s)

$\boxed{\sf{\bold{t = \sqrt{\frac{2 \cdot h}{g}}}}}$

The equation for calculating the velocity (v)

$\boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}}$

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 :

$\sf{h = \frac{1}{2} \cdot g \cdot t^2}$

$\sf{h = \frac{1}{2} \cdot 9.81 \cdot 5^2}$

$\sf{h = \frac{245.25}{2}}$

$\boxed{\sf{h = 122.625 \: m \approx 123 \: m}}$

Conclusion

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

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