The acceleration of gravity on the surface of Mars at the equator is 3.699 m/s2. How long does it take for a rock dropped from a height of 1.045 m to hit the surface

Hi ! This question will direct you to the "Free Fall Motion". Free fall motion is a state of falling objects that are not preceded by any initial velocity. In other words, an object that is in free fall is an object that is "dropped" not "thrown". 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 = 3.699 m/s²
h = the height = 1.045 >> 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 3.699 \times 1.045}} [/tex]

[tex] \sf{v \approx \sqrt{7.731}} [/tex]

[tex] \boxed{\sf{v \approx 2.78 \: m/s}} [/tex]

Conclusion

So, the speed when it (the rock) hit the ground of Mars approximately 2.78 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