A body falling from a high Minaret travels $40 ext{ m}$ in the last $2 ext{ s}$ of its fall to the ground.

Question

A body falling from a high Minaret travels $40 	ext{ m}$ in the last $2 	ext{ s}$ of its fall to the ground. The height of the Minaret in meters is: (take $g=10 	ext{ m/s}^2$)

Accepted Answer

1. **Define Variables:** Let the total height of the Minaret be $H$ and the total time taken to reach the ground be $t$. The body falls from rest, so initial velocity $u = 0$.
2. **Total Distance Equation:** Using the second equation of motion $s = ut + \frac{1}{2}at^2$ with $a = g = 10 	ext{ m/s}^2$:
   $$ H = \frac{1}{2}gt^2 = 5t^2 \quad \dots(1) $$
3. **Distance Before Last 2 Seconds:** The distance covered in $(t - 2)$ seconds is:
   $$ H' = \frac{1}{2}g(t - 2)^2 = 5(t - 2)^2 $$
4. **Apply Condition:** The distance traveled in the last 2 seconds is the difference between the total height and the height covered in $(t-2)$ seconds.
   $$ H - H' = 40 $$
   $$ 5t^2 - 5(t - 2)^2 = 40 $$
   $$ 5[t^2 - (t^2 - 4t + 4)] = 40 $$
   $$ 5(4t - 4) = 40 $$
   $$ 20(t - 1) = 40 \implies t - 1 = 2 \implies t = 3 	ext{ s} $$
5. **Calculate Height:** Substitute $t = 3 	ext{ s}$ back into equation (1):
   $$ H = 5(3)^2 = 5(9) = 45 	ext{ m} $$ .

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