Question

A ball released from the top of a tower travels $\frac{11}{36}$ of the height of the tower in the last second of its journey. The height of the tower is (Take $\left.g=10 \,m \,s ^{-2}\right)$

• A. 11 m
• B. 36 m
• C. 47 m
• D. 180 m

Answer 1

Answer: (D)

$ \frac{11 h}{36}=\frac{10}{2}(2 n-1)$
$\frac{11}{36} \times \frac{1}{2} \times 10 n^{2}=\frac{10}{2}(2 n-1)$ or
$ 2 n-1=\frac{11}{36} n^{2}$
or $ 11 n^{2}=72 n-36$ or $11 n^{2}-72 n+36=0$
or $11 n^{2}-66 n-6 n+36=0$
or $ 11 n(n-6)-6(n-6)=0$ or $(11 n-6)(n-6)=0$
or $ n = 6$ [Rejecting fractional value]
$\therefore h=\frac{1}{2} \times 10 \times 6 \times 6 m =180 \,m$