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].
