Q. 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 $g = 10 = ms^{- 2}$)

A) 11m
B) 36m
C) 47m
D) 180m
Explanation

(d) $\frac{11h}{36} = \frac{9.8}{2}(2n - 1)$
or $\frac{11}{36} \times \frac{1}{2} \times 9.8n^{2} = \frac{9.8}{2}(2n - 1)$
or $2n - 1 = \frac{11}{36}n^{2}$ or $11n^{2} = 72n - 36$
or $11n^{2} - 72n + 36 = 0$
or $11n^{2} - 66n - 6n + 36 = 0$
or $11n(n - 6) - 6(n - 6) = 0$
or $(11n - 6)(n - 6) = 0$
$\Rightarrow \ \ n = 6\ $(Rejecting fractional value)
$h = \frac{1}{2} \times 10 \times 6 \times 6$m=180m

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.