Question

if a freely falling body travels in the last second a distance equal to the distance travelled by it in first 3 seconds . find the time of its travel

Answer 1

Dear student,

The distance covered in first 3 seconds is
$$\begin{gathered} S=\frac{1}{2}g{t}^2 \\ S=\frac{1}{2}\times 9.8\times {\left(3\right)}^2 \\ S=45m \end{gathered}$$

If the ball takes n seconds to to fall down,the distance covered in nth second is
$$\begin{gathered} S=\frac{1}{2}g\left(2n-1\right) \\ S=10n-5 \end{gathered}$$

As distance covered in last second is equal to distance covered in first 3 sec

$$\begin{gathered} 10n-5=45 \\ n=5 \end{gathered}$$

The total height is $h=\frac{1}{2}g{t}^2=\frac{10}{2}\times 25=125m$

The time of its travel is

$$\begin{gathered} t=\sqrt{\frac{2h}{g}} \\ t=\sqrt{\frac{2\times 125}{10}} \\ t=5sec \end{gathered}$$

Regards.