The aim of this question is to understand how to **implement** and **apply** kinematic **equations of motion**.

**Kinematics** is the branch of physics dealing with** objects in motion**. Whenever a body moves in **a straight line,** then the **equations of motion** can be described by the **following formulae:**

\[ v_{ f } \ = \ v_{ i } + a t \]

\[ S = v_{i} t + \dfrac{ 1 }{ 2 } a t^2 \]

\[ v_{ f }^2 \ = \ v_{ i }^2 + 2 a S \]

For the **vertical upward motion:**

\[ v_{ f } \ = \ 0, \ and \ a \ = \ -9.8 \]

In case of **vertical downward motion:**

\[ v_{ i } \ = \ 0, \ and \ a \ = \ 9.8 \]

Where $ v_{ f } $ and $ v_{ i } $ are the final and initial **speed**, $ S $ is the **distance covered**, and $ a $ is the **acceleration.**

## Expert Answer

The given motion can be **divided into two parts**, vertically **upward** motion and vertically **downward** motion.

For the **vertically upward motion:**

\[ v_i \ = \ 8.20 \ m/s \]

\[ v_f \ = \ 0 \ m/s \]

\[ a \ = \ -g \ = \ 9.8 \ m/s^{ 2 } \]

From the **first equation of motion:**

\[ v_{ f } \ = \ v_{ i } + a t \]

\[ \Rightarrow t \ = \ \dfrac{ v_{ f } \ – v_{ i } }{ a } … \ … \ … \ ( 1 ) \]

**Substituting values:**

\[ t \ = \ \dfrac{ 0 \ – 20 }{ -9.8 } \]

\[ \Rightarrow t \ = \ \dfrac{ -20 }{ -9.8 } \]

\[ \Rightarrow t \ = \ 2.04 \ s \]

Since the body has the **same acceleration** and has to cover the **same distance** during the **vertically downward motion, **it will elapse the **same amount of time** as the vertically upward motion. So:

\[ t_{ total } \ = \ 2 \times t \ = \ 4.08 \ s \]

## Numerical Results

\[ t_{ total } \ = \ 4.08 \ s \]

## Example

Calculate the **distance covered** by the bowling pin **during the upward motion**.

For the **vertically upward motion:**

\[ v_i \ = \ 8.20 \ m/s \]

\[ v_f \ = \ 0 \ m/s \]

\[ a \ = \ -g \ = \ 9.8 \ m/s^{ 2 } \]

From the **3rd equation of motion:**

\[ v_{ f }^2 \ = \ v_{ i }^2 + 2 a S \]

\[ \Rightarrow S \ = \ \dfrac{ v_{ f }^2 \ – \ v_{ i }^2 }{ 2 a } \]

**Substituting values:**

\[ \Rightarrow S \ = \ \dfrac{ ( 0 )^2 \ – \ ( 8.20 )^2 }{ 2 ( -9.8 ) } \]

\[ \Rightarrow S \ = \ \dfrac{ – 67.24 }{ – 19.6 } \]

\[ \Rightarrow S \ = \ 3.43 \ m \]