Question

A stone initially at rest is let to fall at time t = 0 from a height of 125m from the earth's surface. What is the distance traveled by the stone during the last second of its journey ?

Answer 1

your answer is 0. mate hope something. mistakes in .......

Answer 2

The distance traveled by the stone during the last second is $S_{l}=45m$ .

Given,

A stone with,

Initial speed, $u=0m/s$

Initial time, $t=0s$

Total distance, $s=125m$

To find,

Distance traveled by the stone during the last second of its journey.

Solution,

Using Newton's second equation of motion,

$S=ut+\frac{1}{2}at^{2}$

Let us suppose the acceleration due to gravity, $g=10m/s$.

So,

$S=ut+\frac{1}{2}gt^{2}$

For the last second,

$S= 125m$

So,

$125=0t$×$\frac{1}{2}(10)t^{2}$

⇒$125=5t^{2}$

⇒$t^{2}=\frac{125}{5}$

⇒$t=5sec$

To find the distance traveled by the stone during the last second, we need to find the distance at  $t=4sec.$

⇒$S_{4}=0+\frac{1}{2}(10)(4)^{2}$

⇒$S_{4}=5(16)$

⇒$S_{4}=80m$

Therefore,

Distance traveled at last second,

$S_{l}=S-S_{4}$

⇒$S_{l}=125m-80m$
⇒$S_{l}=45m$

Therefore,

The distance traveled by the stone during the last second of its journey is $S_{l}=45m$ .