Understanding Object Fall from Rest

This problem involves an object undergoing free fall, starting from rest. We are given information about the distance covered in the last second of its fall and asked to find the total height of its fall.

Setting up the Equations for Height of its Fall

Let the total height of its fall be \(H\) and the total time taken for the fall be \(T\). The object starts from rest, meaning its initial velocity \(u = 0\). The acceleration is due to gravity, denoted by \(g\). We use the kinematic equation relating distance, initial velocity, time, and acceleration:

\(s = ut + \frac{1}{2}at^2\)

For the total fall, \(s = H\), \(u = 0\), \(a = g\), and \(t = T\). So, the total height of its fall is given by:

\(H = 0 \cdot T + \frac{1}{2}gT^2\)

\(H = \frac{1}{2}gT^2\) (Equation 1)

Distance Covered in the Last Second of Fall

The problem states the object travels half its total path in the last second. This means the distance covered in the time \((T-1)\) seconds is the other half of the total height, i.e., \(\frac{H}{2}\).

The distance covered in the first \((T-1)\) seconds, starting from rest, is:

\(H_{T-1} = \frac{1}{2}g(T-1)^2\)

According to the problem:

\(H_{T-1} = \frac{H}{2}\)

So, we have the equation:

\(\frac{1}{2}g(T-1)^2 = \frac{H}{2}\) (Equation 2)

Solving for Total Time (T)

Now we substitute the expression for \(H\) from Equation 1 into Equation 2:

\(\frac{1}{2}g(T-1)^2 = \frac{1}{2} \left(\frac{1}{2}gT^2\right)\)

\(\frac{1}{2}g(T-1)^2 = \frac{1}{4}gT^2\)

Assuming \(g \neq 0\), we can divide both sides by \(\frac{1}{2}g\):

\((T-1)^2 = \frac{1}{2}T^2\)

Expand the left side and move all terms to one side:

\(T^2 - 2T + 1 = \frac{1}{2}T^2\)

\(T^2 - \frac{1}{2}T^2 - 2T + 1 = 0\)

\(\frac{1}{2}T^2 - 2T + 1 = 0\)

Multiply by 2 to get integer coefficients:

\(T^2 - 4T + 2 = 0\)

This is a quadratic equation in \(T\). We use the quadratic formula \(T = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) with \(a=1, b=-4, c=2\):

\(T = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(2)}}{2(1)}\)

\(T = \frac{4 \pm \sqrt{16 - 8}}{2}\)

\(T = \frac{4 \pm \sqrt{8}}{2}\)

\(T = \frac{4 \pm 2\sqrt{2}}{2}\)

\(T = 2 \pm \sqrt{2}\)

We get two possible values for \(T\): \(T_1 = 2 + \sqrt{2}\) and \(T_2 = 2 - \sqrt{2}\). The problem states the object travels half the distance in the *last second*. This implies the total time of fall \(T\) must be greater than 1 second.

Let's approximate the values:

\(\sqrt{2} \approx 1.414\)

\(T_1 \approx 2 + 1.414 = 3.414\) seconds (This is greater than 1 second, so it is a valid time)

\(T_2 \approx 2 - 1.414 = 0.586\) seconds (This is less than 1 second, so it is not a valid time in this context)

So, the total time of fall is \(T = 2 + \sqrt{2}\) seconds.

Calculating the Total Height of its Fall

Now we use the valid value of \(T\) and Equation 1 to find the total height of its fall. We use the standard value for acceleration due to gravity, \(g \approx 9.8 \, m/s^2\).

\(H = \frac{1}{2}gT^2\)

\(H = \frac{1}{2}(9.8)(2 + \sqrt{2})^2\)

\(H = 4.9 ( (2)^2 + 2(2)\sqrt{2} + (\sqrt{2})^2 )\)

\(H = 4.9 (4 + 4\sqrt{2} + 2)\)

\(H = 4.9 (6 + 4\sqrt{2})\)

Using \(\sqrt{2} \approx 1.414\):

\(H \approx 4.9 (6 + 4 \times 1.414)\)

\(H \approx 4.9 (6 + 5.656)\)

\(H \approx 4.9 (11.656)\)

\(H \approx 57.1144 \, m\)

This calculated value for the height of its fall is approximately 57.1144 meters, which is closest to 57 m among the given options.