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

Question

Dr. Abubacker · Accepted Answer

(b) Let height of minaret is $H$ and body take time $T$ to fall from top to bottom
$H = \frac{1}{2}gT^{2}$ …(i)
In last $2\ sec$ body travels distance of $40\ m$ so in $(T - 2)\ sec$ distance travelled $= (H - 40)m$
$(H - 40) = \frac{1}{2}g(T - 2)^{2}$ …(ii)
By solving (i) and (ii), $T = 3\ sec$ and $H = 45m$

Q. A body falling from a high Minaret travels $40\ m$ in the last 2 seconds of its fall to ground. Height of Minaret in meters is (take $g = 10\ m/s^{- 2}$)

Explanation

Reviewed by Dr. Abubacker

Want more practice?