This question aims to deploy the **kinematic** **e****quations** commonly known as the **equations of motion**. It covers a special case of 2-D motion known as the **projectileÂ motion**.

The** distance** $ ( S ) $ covered in unit amount of time time $ ( t ) $ is known as speed $ ( v ) $. It is mathematically defined as:

\[ v \ = \ \dfrac{ S }{ t } \]

The **straight line equations****Â of motion** can be described by the following formula:

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

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

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

In case of **vertical upward motion**:

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

In case of **vertical downward motion**:

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

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

We canÂ use aÂ **combination of**Â the above **constraints and equations**Â to solve the given problem.

In the **context of the given question,** the **animal is jumping at an angle** of 45 degrees so it will not follow a perfectly vertical path. Rather, it will perform a **projectile motion**. For the case of projectile motion, the **maximum height** can be calculated using the following **mathematical formula.**

The most important parameters during theÂ **flight of aÂ projectile**Â are itsÂ **range**,Â **time of flight**, andÂ **maximum height**.

TheÂ **range of aÂ projectile**Â is given by the following formula:

\[ R \ = \ \dfrac{ v_i^2 \ sin ( 2 \theta ) }{ g } \]

The**Â time of flight**Â of aÂ projectileÂ is given by the following formula:

\[ t \ = \ \dfrac{ 2 v_i \ sin \theta }{ g } \]

TheÂ **maximum height**Â of aÂ projectileÂ is given by the following formula:

\[ h \ = \ \dfrac{ v_i^2 \ sin^2 \theta }{ 2 g } \]

## Expert Answer

For the **projectile motion:**

\[ h \ = \ \dfrac{ v_i^2 \ sin^2 \theta }{ 2 g } \]

**Re-arranging** this equation:

\[ v_i^2 \ = \ \dfrac{ 2 g h }{ sin^2 \theta } \]

\[ \Rightarrow v_i \ = \ \sqrt{ \dfrac{ 2 g h }{ sin^2 \theta } } \]

\[ \Rightarrow v_i \ = \ \dfrac{ \sqrt{ 2 g h } }{ sin \theta } … \ … \ … \ ( 1 ) \]

**Substituting values:**

\[ v_i \ = \ \dfrac{ \sqrt{ 2 ( 9.8 ) ( 3.7 ) } }{ sin ( 45^{ \circ } ) } \]

\[ \Rightarrow v_i \ = \ \dfrac{ \sqrt{ 72.52 } }{ 0.707 } \]

\[ \Rightarrow v_i \ = \ 12.04 \ m/s \]

## Numerical Result

\[ v_i \ = \ 12.04 \ m/s \]

## Example

In the** same scenario** given above, calculate the **initial speed required** to achieve a **height of 1 m.**

Using the same formula of height in **equation (1):**

\[ v_i \ = \ \dfrac{ \sqrt{ 2 g h } }{ sin \theta } \]

**Substituting values:**

\[ v_i \ = \ \dfrac{ \sqrt{ 2 ( 9.8 ) ( 1 ) } }{ sin ( 45^{ \circ } ) } \]

\[ \Rightarrow v_i \ = \ \dfrac{ \sqrt{ 19.60 } }{ 0.707 } \]

\[ \Rightarrow v_i \ = \ 6.26 \ m/s \]