Q58P - University Physics with Modern Physics

Question

A brick is dropped from the roof of a tall building. After it has been falling for a few seconds, it falls 40.0 m in a 1.00-s time interval. What distance will it fall during the next 1.00 s? Ignore air resistance.

Step-by-Step Solution

Answer

the distance the brick will fall in the next 1.00 s is 49.8 m.

Step 1: Identification of the given data

The given data can be expressed below as:

• The brick falls to about 40.0 m.
• The brick falls at a time interval of 1.00 s.
• The air resistance is 0 for the brick.

Step 2: Significance of the Newton’s first law for the brick

This law states that an object will continue to move in uniform velocity if the object is not acted by an external force.

The equation of the displacement of the brick gives the distance the brick will fall in the next 1.00 s.

Step 3: Determination of the distance the brick will fall

From the equation of displacement along a straight line, the displacement of the brick can be expressed as:

$y-y_0=V_{0}t+\frac{1}{2}g{t}^2$

Here, $y-y_0$ is the displacement of the brick that is 40 m, $v_0$ is the initial velocity, t is the time taken for the brick that is 1 s and the value of g is $9.8\hspace{0.33em}m/s^2$.

Substituting the values in the above equation, we get-

$$\begin{gathered} 40=v_0+\frac{1}{2}\times 9.8\mathrm{m}/{\mathrm{s}}^2\times (1{)}^2 \\ {v}_0=35.1\frac{\mathrm{m}}{\mathrm{s}} \end{gathered}$$

As the brick will drop in the next 1 s, hence, the displacement in t =2s will be-

$$\begin{gathered} y-y_0=35.1\mathrm{m}/\mathrm{s}\times 2+4.9\mathrm{m}/{\mathrm{s}}^2\times (2\mathrm{s}{)}^2 \\ =89.8\mathrm{m} \end{gathered}$$

So, the distance moved in the final 1-s interval is:

$$\begin{gathered} \mathrm{d}=89.8 \mathrm{m}-40 \mathrm{m} \\ =49.8 \mathrm{m} \end{gathered}$$

Thus, the distance the brick will fall in the next 1.00 s is 49.8 m.