MCQ
A body falling from a high Minaret travels $40m$ in the last $2$ seconds of its fall to ground. Height of Minaret in metres is: $($take $g = 10\ m/ s^2)$
- A$60$
- ✓$45$
- C$80$
- D$50$
✓
Answer
Correct option: B.
$45$
Taking the height of minaret is $H$ and time taken by body to fall from top to bottom be $T.$
$\therefore\text{H}=\big(\frac{1}{2}\big)\text{g}\text{T}^2.....(1)$
In last two second body travels a distance of $40m,$
hence ein $(T - 2)\sec$ body will travel $(H - 40)m.$
$(\text{H}-40)=\big(\frac{1}{2}\big)\text{g}(\text{T}-2)^2.....(2)$
$\therefore$ solving $(1)$ and $(2),$
$T = 3\sec, H = 45m$
$\therefore\text{H}=\big(\frac{1}{2}\big)\text{g}\text{T}^2.....(1)$
In last two second body travels a distance of $40m,$
hence ein $(T - 2)\sec$ body will travel $(H - 40)m.$
$(\text{H}-40)=\big(\frac{1}{2}\big)\text{g}(\text{T}-2)^2.....(2)$
$\therefore$ solving $(1)$ and $(2),$
$T = 3\sec, H = 45m$
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