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

Step 1: Know the free fall rule.
The body starts from rest and falls only under gravity. So the distance fallen from the start in time $t$ is \[ s = \tfrac12 g t^2 \] To find how far it moves during a chosen slice of time, we subtract the distance at the start of that slice from the distance at the end.

Step 2: Find total fall up to each needed time.
With $g = 10$, we compute the fall for the times we care about. \[ s(0.5)=1.25,\; s(1)=5,\; s(1.5)=11.25 \] \[ s(2)=20,\; s(2.5)=31.25,\; s(3)=45 \] All values are in metres.
Step 3: Second half-second slice.
This slice runs from $0.5$ s to $1$ s. The distance moved in it is \[ 5 - 1.25 = 3.75\,\text{m} \]
Step 4: Fourth half-second slice.
This runs from $1.5$ s to $2$ s. \[ 20 - 11.25 = 8.75\,\text{m} \]
Step 5: Sixth half-second slice.
This runs from $2.5$ s to $3$ s. \[ 45 - 31.25 = 13.75\,\text{m} \]
Step 6: Add the three required slices.
The question asks for the total of these three displacements. \[ 3.75 + 8.75 + 13.75 = 26.25\,\text{m} \] \[ \boxed{26.25\,\text{m}} \]