When an object is dropped, the number of feet N that it falls is given

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Hi Gmatprep550 ,
I can understand where you are coming from.

My first thought was the same as yours:
But velocity increases rapidly!

In a way, because we do not import any outside knowledge,
these questions are similar to special symbol questions.

The formula may or may not be accurate . . .
and it is true only for the question at hand.

I shifted gears.
What are they testing? I asked myself.
What are we given? A total time and a partial time.
In what context?
We are given a formula for distance.

time . . . distance . . . and rate.

• In other words, we can start with something that we know
Use \(rt=D\)

If we focus only on \(rt=D\) (ignore the formula for a moment), then we
can write something such as:

Distance total? t = 5 seconds: \(r * 5 = D_{total}\)
Distance partial? t = 2 seconds: \(r * 2 = D_{partial}\)

\(r\) is constant. \(D\) is fixed.
Rate is constant (inane, but the prompt does not say otherwise)
Aha! Time taken is proportional to distance traveled.
Factor the constant \(r\) out.
\(\frac{D_{partial}}{D_{total}}=\frac{2secs}{5secs} = \frac{2}{5}\)

In 2 seconds, the object will travel \(\frac{2}{5}\) of the distance that it traveled in 5 seconds.

• Formula
\(N=\frac{1}{2} gt^2\),
N = number of feet = distance, D

Find \(D_{total}\), i.e., the distance that the object travels in 5 seconds

Plug 5 into the "special" formula.
I'm changing N to D. It's easier for me.

• TOTAL distance
\(D_{total}=\frac{1}{2} gt^2\)
\(D_{total}=\frac{1}{2} * 32.2 * 5^2\)
\(D_{total}= 402.5=\) total D

• Distance in 2 seconds?
-- The distance that the object travels in 2 seconds is \(\frac{2}{5}\) of 402.5

\(D_{2secs} = \frac{2}{5} * 402.5\)
\(D_{2secs} = 161\)

Answer C

Hope that helps. Ask again if it does not.

(For the record, physics and I were Not Friends ,
except for quantum mechanics, but I learned the latter in chemistry.)