Question

A ball is dropped from a height of 80m. What is the ratio of the distances traveled by it during the first and the last second of its motion.
( answer should come 1:7 )

Answer 1

$S = u t + 1/2 g t^2 \\ \\ t = time\ of\ flight\ = \sqrt{\frac{2 * 80}{10}} = 4 sec \\ \\ S_1 = u +(n-1/2) g = 0 + 1/2 g \\ \\ S_4 = 0+ (4-1/2) g = 7/2 g \\ Ratio = \frac{\frac{1g}{2}}{ \frac{7g}{2}} \\ \\ 1/7$