Question

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

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:

$\boxed{\sf{\bold{v = \sqrt{2 \times g \times h}}}}$ ... (i)

$\boxed{\sf{\bold{v = \sqrt{2 \times g \times (h_1 - h_2)}}}}$ ... (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)
• $\sf{h_1}$ = initial high (m)
• $\sf{h_2}$ = 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/s²
• h = 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 :

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

$\sf{v = \sqrt{2 \times 9.8 \times 10}}$

$\sf{v = \sqrt{196}}$

$\boxed{\sf{v = 14 \: m/s}}$

Conclusion

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

See More

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