Welcome to Westonci.ca, your ultimate destination for finding answers to a wide range of questions from experts. Get expert answers to your questions quickly and accurately from our dedicated community of professionals. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
Certainly! Let's solve this problem step-by-step using the principles of physics, specifically the equations of motion under gravity.
Given:
- The height ([tex]\(h\)[/tex]) from which the object is dropped is [tex]\(1000\)[/tex] feet.
- The acceleration due to gravity ([tex]\(g\)[/tex]) is [tex]\(32.2 \, \text{ft/s}^2\)[/tex].
We will use the kinematic equation for objects in free fall:
[tex]\[ v^2 = u^2 + 2gh \][/tex]
Here:
- [tex]\(v\)[/tex] is the final velocity of the object.
- [tex]\(u\)[/tex] is the initial velocity (which is [tex]\(0\)[/tex] since the object is dropped).
- [tex]\(g\)[/tex] is the acceleration due to gravity.
- [tex]\(h\)[/tex] is the height from which the object is dropped.
Since the object is dropped from rest:
[tex]\[ u = 0 \][/tex]
Substituting this into the equation, we get:
[tex]\[ v^2 = 0^2 + 2gh \][/tex]
[tex]\[ v^2 = 2gh \][/tex]
Now, let's plug in the given values:
[tex]\[ v^2 = 2 \cdot 32.2 \cdot 1000 \][/tex]
So:
[tex]\[ v^2 = 64400 \][/tex]
To find [tex]\(v\)[/tex], we take the square root of both sides:
[tex]\[ v = \sqrt{64400} \][/tex]
The value of [tex]\(\sqrt{64400}\)[/tex] is approximately [tex]\(253.77 \, \text{ft/s}\)[/tex].
Therefore, the speed of the object when it reaches the ground is approximately [tex]\(\boxed{253.77 \, \text{ft/s}}\)[/tex].
Given:
- The height ([tex]\(h\)[/tex]) from which the object is dropped is [tex]\(1000\)[/tex] feet.
- The acceleration due to gravity ([tex]\(g\)[/tex]) is [tex]\(32.2 \, \text{ft/s}^2\)[/tex].
We will use the kinematic equation for objects in free fall:
[tex]\[ v^2 = u^2 + 2gh \][/tex]
Here:
- [tex]\(v\)[/tex] is the final velocity of the object.
- [tex]\(u\)[/tex] is the initial velocity (which is [tex]\(0\)[/tex] since the object is dropped).
- [tex]\(g\)[/tex] is the acceleration due to gravity.
- [tex]\(h\)[/tex] is the height from which the object is dropped.
Since the object is dropped from rest:
[tex]\[ u = 0 \][/tex]
Substituting this into the equation, we get:
[tex]\[ v^2 = 0^2 + 2gh \][/tex]
[tex]\[ v^2 = 2gh \][/tex]
Now, let's plug in the given values:
[tex]\[ v^2 = 2 \cdot 32.2 \cdot 1000 \][/tex]
So:
[tex]\[ v^2 = 64400 \][/tex]
To find [tex]\(v\)[/tex], we take the square root of both sides:
[tex]\[ v = \sqrt{64400} \][/tex]
The value of [tex]\(\sqrt{64400}\)[/tex] is approximately [tex]\(253.77 \, \text{ft/s}\)[/tex].
Therefore, the speed of the object when it reaches the ground is approximately [tex]\(\boxed{253.77 \, \text{ft/s}}\)[/tex].
We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.