Question

A body is falling freely under gravity from a height of \(200\,\text{m}\). The total displacement of the body during the second half-second, fourth half-second and sixth half-second of its motion is (Acceleration due to gravity \(=10\,\text{m s}^{-2}\)):

• A. \(26.25\,\text{m}\)
• B. \(32.50\,\text{m}\)
• C. \(37.25\,\text{m}\)
• D. \(42.25\,\text{m}\)

Hint:

For uniformly accelerated motion: \[ s_n = u + \frac{a}{2}(2n-1) \] gives displacement in the \(n^{th}\) second. This shortcut is very useful in free-fall problems.

Answer 1

The Correct Option is A

Concept: For a body falling freely from rest under gravity: \[ s = \frac{1}{2}gt^2 \] Displacement during a particular interval is: \[ \Delta s = s_2 - s_1 \] Each half-second interval has duration: \[ 0.5\,\text{s} \]

Step 1: Find displacement during the second half-second. The second half-second corresponds to interval: \[ 0.5\,\text{s} \text{ to } 1\,\text{s} \] Distance fallen in \(1\,\text{s}\): \[ s_1 = \frac{1}{2}(10)(1)^2 = 5\,\text{m} \] Distance fallen in \(0.5\,\text{s}\): \[ s_{0.5} = \frac{1}{2}(10)(0.5)^2 \] \[ =5 \times 0.25 \] \[ =1.25\,\text{m} \] Hence displacement during second half-second: \[ 5 - 1.25 = 3.75\,\text{m} \]

Step 2: Find displacement during fourth half-second. Fourth half-second corresponds to: \[ 1.5\,\text{s} \text{ to } 2\,\text{s} \] Distance fallen in \(2\,\text{s}\): \[ s_2 = \frac{1}{2}(10)(2)^2 = 20\,\text{m} \] Distance fallen in \(1.5\,\text{s}\): \[ s_{1.5} = \frac{1}{2}(10)(1.5)^2 \] \[ =5(2.25) \] \[ =11.25\,\text{m} \] Displacement: \[ 20 - 11.25 = 8.75\,\text{m} \]

Step 3: Find displacement during sixth half-second. Sixth half-second corresponds to: \[ 2.5\,\text{s} \text{ to } 3\,\text{s} \] Distance fallen in \(3\,\text{s}\): \[ s_3 = \frac{1}{2}(10)(3)^2 \] \[ =45\,\text{m} \] Distance fallen in \(2.5\,\text{s}\): \[ s_{2.5} = \frac{1}{2}(10)(2.5)^2 \] \[ =5(6.25) \] \[ =31.25\,\text{m} \] Displacement: \[ 45 - 31.25 = 13.75\,\text{m} \]

Step 4: Add all required displacements. \[ 3.75 + 8.75 + 13.75 \] \[ =26.25\,\text{m} \] Therefore, \[ \boxed{26.25\,\text{m}} \]