Question

A particle is dropped from the top of a tower. During its motion it covers $\dfrac{9}{{25}}$ ​ part of the height of the tower in the last 1 seconds. Then find the height of the tower.

Answer 1

HINT: Proceed the solution of this question, first writing one of the three equation of linear motion which is of distance ${\text{s = }}\dfrac{1}{2}{\text{a}}{{\text{t}}^2}$ as it dropped case so we can assume initial velocity u equal to zero. The last second is given in terms of total distance. Hence by writing distance equation twice, for last second and for remaining time, common terms will cancel and we will get the required answer.

Complete step-by-step answer:
Ball travels $\dfrac{9}{{25}}$​ of the height (say $\space x$) in the t second (last second)

⇒Hence it travelled $\dfrac{{16}}{{25}}x$ in (t−1) seconds.
and the total distance in t seconds.

The ball is dropped hence initial velocity is zero and gravity is $9.8{\text{ }}\dfrac{{\text{m}}}{{{s^2}}}$

Hence distance travelled is

⇒${\text{s = }}\dfrac{1}{2}{\text{a}}{{\text{t}}^2} = \dfrac{1}{2}{\text{g}}{{\text{t}}^2}$
⇒$x = \dfrac{1}{2}{\text{g}}{{\text{t}}^2}$ ……(1)
⇒$\dfrac{{16}}{{25}}x = \dfrac{1}{2}{\text{g}}{\left( {{\text{t - 1}}} \right)^2}$ …..(2)
Divide the equations (2) by (1)
$ \Rightarrow \dfrac{{16}}{{25}} = \dfrac{{{{\left( {{\text{t - 1}}} \right)}^2}}}{{{{\text{t}}^2}}}$
⇒$16{{\text{t}}^2} = 25{\left( {{\text{t - 1}}} \right)^2}$ …..(3)
On taking square root both side
$ \Rightarrow {\text{4t}} = \pm 5\left( {{\text{t - 1}}} \right)$
$ \Rightarrow {\text{4t}} = 5{\text{t - 5}}$
$ \Rightarrow t = 5$

Which gives t = 5 seconds

And taking negative signs which will give negative time which is not possible.

Hence height $x = \dfrac{1}{2} \times {\text{9}}{\text{.8}} \times {5^2}$ = 122.5 m

Note- In the question of vertical motion, it is advisable to remember some general formulas of linear motion equation which are as

$ \Rightarrow {\text{v = u + at}}$
$ \Rightarrow {\text{s = ut + }}\dfrac{1}{2}{\text{a}}{{\text{t}}^2}$
$ \Rightarrow {{\text{v}}^2} = {u^2} + 2as$

Hence it is vertical motion these linear equation of motion will be converted into vertical motion by changing acceleration a by gravitational constant g

$ \Rightarrow {\text{v = u - gt}}$
$ \Rightarrow {\text{s = ut - }}\dfrac{1}{2}{\text{g}}{{\text{t}}^2}$
$ \Rightarrow {{\text{v}}^2} = {u^2} - 2gh$