Physics College

A rock is dropped at 10 m height at what speed does it hit the ground?

Answers

Answer 1

So, the speed when it (the rock) hit the ground is 14 m/s.

Introduction

Hi ! This question will direct you to the "Free Fall Motion" material. Why is that ? This is because there is a keyword "dropped" which means without initial speed and its named free fall motion. Its little bit of different, when the keyword "thrown", that's not a free fall motion. For the equations of velocity in free fall, follow the following equation:

[tex] \boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}} [/tex] ... (i)

[tex] \boxed{\sf{\bold{v = \sqrt{2 \times g \times (h_1 - h_2)}}}} [/tex] ... (ii)

With the following condition :

v = velocity (m/s)g = acceleration of the gravity (m/s²)h = the height or displacement at vertical line (m)[tex] \sf{h_1} [/tex] = initial high (m)[tex] \sf{h_2} [/tex] = final high (m)

Note :

Use the (i) equation when the object has actually touched the ground (the final position of the object is at the ground).Use the (ii) equation when the object not touched the ground (the final position of the object is > 0 m above the ground).

Problem Solving

We know that :

g = acceleration of the gravity = 9.8 m/s²h = the height = 10 m >> The final position of the object is directly touch the ground.

What was asked :

v = velocity = ... m/s

Step by step :

[tex] \sf{v = \sqrt{2 \times g \times h}} [/tex]

[tex] \sf{v = \sqrt{2 \times 9.8 \times 10}} [/tex]

[tex] \sf{v = \sqrt{196}} [/tex]

[tex] \boxed{\sf{v = 14 \: m/s}} [/tex]

Conclusion

So, the speed when it (the rock) hit the ground is 14 m/s.

See MoreThe speed of the object at a certain height (free fall motion) https://brainly.com/question/26377041The relationship between acceleration and the change in velocity and time in free fall https://brainly.com/question/26486625

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