The aim of this question is to utilize the **equations of motion** for solving 2D **motion-related problems**.

The speed is the **rate of change of distance** **s** with respect to time **t**:

** v = s/t**

If **vf** is the **final speed**, **vi** is the **initial speed**, **a** is the **acceleration** and **s** is the **distance** covered, the **equations of motion** are given by:

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

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

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

For **vertical upward motion**:

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

For **vertical downward motion**:

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

We will use a **combination of** the above c**onstraints and equations** to solve the given problem.

## Expert Answer

Using the **3rd equation of motion** in the vertical direction:

\[ v_{ fy }^2 \ = \ v_{ iy }^2 + 2 a S \]

Substituting values:

\[ ( 0 )^2 \ = \ v_{ iy }^2 + 2 ( -9.8 ) ( 3 ) \]

\[ \Rightarrow 0 \ = \ v_{ iy }^2 \ – \ 58.8 \]

\[ \Rightarrow v_{ iy }^2 \ = \ 58.8 \]

\[ \Rightarrow v_{ iy } \ = \ \sqrt{ 58.8 } \]

\[ \Rightarrow v_{ iy } \ = \ 7.668 m/s \]

Using **second equation of motion**:

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

Substituting values:

\[ 3 \ = \ ( 0 ) t + \dfrac{ 1 }{ 2 } (9.8) t^2 \]

\[ \Rightarrow 3 \ = \ 4.9 t^2 \]

\[ \Rightarrow t \ = \ \sqrt{ \dfrac{ 3 }{ 4.9 } } \]

\[ \Rightarrow t \ = \ 0.782 \ s\]

Using the formula for **speed in horizontal direction**:

\[ v_x \ = \ \dfrac{ 10 }{ 0.782 } = 12.78 \ m/s \]

Calculating the **magnitude of the speed**:

\[ |v| \ = \ \sqrt{ v_x^2 \ + \ v_y^2 } \]

\[ \Rightarrow |v| \ = \ \sqrt{ ( 12.78 )^2 \ + \ ( 7.668 )^2 } \]

\[ \Rightarrow |v| \ = \ 14.9 \ m/s \]

Calculating the **direction of the speed**:

\[ \theta \ = \ tan^{-1} \bigg ( \dfrac{ v_y }{ v_x } \bigg ) \]

\[ \theta \ = \ 36.9^{ \circ } \]

## Numerical Result

\[ v \ = \ 14.9 \ m/s \text{ at } \theta = 36.9^{ \circ } \text{ from ground } \]

## Example

A **man makes a leap** $ 2.0 \ m $ long and $ 0.5 \ m $ high. What is the **speed of the man** just as he leaves the ground?

Using the **3rd equation of motion** in the vertical direction:

\[ v_{ fy }^2 \ = \ v_{ iy }^2 + 2 a S \]

\[ \Rightarrow v_{ iy } \ = \ \sqrt{ -2 a S – v_{ fy }^2 } \]

\[ \Rightarrow v_{ iy } \ = \ \sqrt{ -2 ( -9.8 ) ( 0.5 ) – 0 } \ = \ 9.8 \ m/s \]

Using **second equation of motion**:

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

\[ 0.5 \ = \ ( 0 ) t + \dfrac{ 1 }{ 2 } (9.8) t^2 \]

\[ \Rightarrow t \ = \ \sqrt{ \dfrac{ 0.5 }{ 4.9 } } \ = \ 0.32 \ s \]

Using the formula for **speed in horizontal direction**:

\[ v_x \ = \ \dfrac{ 2 }{ 0.32 } = 6.25 \ m/s \]

Calculating the **magnitude of the speed**:

\[ |v| \ = \ \sqrt{ v_x^2 \ + \ v_y^2 } \ = \ \sqrt{ ( 6.25 )^2 \ + \ ( 9.8 )^2 } \ = \ 11.62 \ m/s \]

Calculating the **direction of the speed**:

\[ \theta \ = \ tan^{-1} \bigg ( \dfrac{ v_y }{ v_x } \bigg ) \ = \ tan^{-1} \bigg ( \dfrac{ 9.8 }{ 6.25 } \bigg ) \ = \ 57.47^{ \circ } \]