Question

A body falls freely from a height of $490m$. Determine the displacement of the body during the last second of its descent.

Answer 1

Freely Falling Bodies:

1. When a body is dropped from a height without any initial velocity given to the body under the influence of gravity of the earth said to be falling freely.
2. The initial velocity of the body is zero.
3. The acceleration applied to the body is equal to the gravity of the earth and its direction is downwards.

Step 1: Given Data

The height from which the body falls, $h=490m$.

The initial velocity of the body, $u=0m{s}^{-1}$ because the body is falling freely.

(Note: The acceleration due to gravity on earth is assumed as $g=9.8m{s}^{-2}$ and the downward direction is considered positive.)

Step 2: Determine the time taken by the body to reach the ground.

Let the time taken for the body to reach the ground be $t$.

As we know the second equation of motion is $\mathit{S}\mathbf{}\mathbf{=}\mathbf{}\mathit{u}\mathit{t}\mathbf{}\mathbf{+}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}\mathit{a}{\mathit{t}}^{\mathbf{2}}$

On putting the given values in the above equation we get,

$\Rightarrow h=ut+\frac{1}{2}g{t}^2$

$\Rightarrow 490=0+\frac{1}{2}\times 9.8\times t^2$

$\Rightarrow t^2=\frac{490}{4.9}$

$\Rightarrow t^2=100$

$\Rightarrow t=\sqrt{100}$

$\Rightarrow t=10s$

Step 3: Calculate the displacement during the last second of its descent.

The formula to calculate the displacement in the last second of the free fall is ${\mathbf{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathbf{g}\left[{t_n}^2-{\left(t_n-1\right)}^2\right]$.

Here, $t_n$ is the time taken by the body to reach the ground and $t_n=t=10s$ (Calculated in step 2).

On substituting the values we get,

$\Rightarrow S_n=\frac{1}{2}\times 9.8\times \left({10}^2-9^2\right)$

$\Rightarrow S_n=4.9\times 19$

$\Rightarrow S_n=93.1m$

Final Answer:

The displacement of the body during the last second of its descent is $\mathbf{93}\mathbf{.}\mathbf{1}\mathbf{}\mathit{m}$