Set 4 Problem number 14

• Problem

• Solution 

• Generalized Solution

Problem

During an observation, an object released from rest undergoes a free fall of 4 meters. What is the minimum possible time of observation?

Solution

In the vicinity of Earth's surface, the object will accelerate at 9.8 m/s/s.

• If we were given the time duration and acceleration we could reason this problem out in two steps; but with the given information there is no easy way to reason out the time.
• So we use the formula `ds = v0 * `dt + .5 a `dt ^ 2, with v0=0, a=9.8 m/s/s and `ds= 4 m.
• Since v0 = 0 the formula becomes `ds = .5 a `dt^2.
• Solving for `dt we obtain `dt = `sqrt{2 `ds / a}

• Note that   -`sqrt{2 `ds / a} is also a possible solution, but in this situation we reject the negative `dt as making no sense.

• Substituting we obtain

• `dt = `sqrt(2 * 4 m / (9.8 m/s^2) )  = .903 s.

Generalized Solution

Starting from rest and accelerating at constant rate g we have

• `ds = v0 `dt + .5 a `dt^2 = .5 a `dt^2, since v0 = 0.

We solve `ds = .5 a `dt^2 for `dt and choose the positive value of `dt:

• `dt = `sqrt ( 2 `ds / a ).