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An object is dropped from a platform 1000 feet high. Ignoring wind resistance, what will its speed be when it reaches the ground?

A. 2000 ft/s
B. 44 ft/s
C. 80 ft/s
D. 6400 ft/s


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].