Question

A stone is dropped from the top of a tower of hight 245m. What is the distance covered by stone during it's last one second of motion?

Answer 1

Solution :

Given :-

▪ A stone is dropped from the top of a tower of height 245m.

To Find :-

▪ Distance covered by the stone during it's last one second of motion.

Concept :-

First, we have to find out total time taken by stone to come to the ground after that we can calculate distance covered by stone in last second.

Formula of distance covered by moving body in nth second is given by

$\boxed{\bf{\pink{s_{n}=u+\dfrac{a}{2}(2n-1)}}}$

Calculation :-

• Calculation of time

$\implies\bf\:h=ut+\dfrac{1}{2}gt^2\\ \\ \implies\sf\:h=(0\times t)\dfrac{1}{2}gt^2\\ \\ \implies\bf\purple{t=\sqrt{\dfrac{2h}{g}}}\\ \\ \implies\sf\:t=\sqrt{\dfrac{2\times 245}{10}}\\ \\ \implies\sf\:t=\sqrt{49}\\ \\ \implies\underline{\bf{t=7\:sec}}$

• Calculation of distance

$\mapsto\sf\:s_n=0+\dfrac{10}{2}[(2\times 7)-1]\\ \\ \mapsto\sf\:s_n=5(14-1)\\ \\ \mapsto\sf\:s_n=5(13)\\ \\ \mapsto\boxed{\orange{\bf{s_n=65\:m}}}$