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The best leaper in the animal kingdom is the puma, which can jump to a height of 3.7 m when leaving the ground at an angle of 45 degrees. With what speed must the animal leave the ground to reach that height?

This question aims to deploy the kinematic equations commonly known as the equations of motion. It covers a special case of 2-D motion known as the projectile 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 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 final  and initial speed, $ S $ is the distance covered, and $ a $ is the acceleration.

We can use a combination of the above constraints and equations 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 the flight of a projectile are its rangetime of flight, and maximum height.

The range of a projectile is given by the following formula:

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

The time of flight of a projectile is given by the following formula:

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

The maximum height of a projectile 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:

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

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

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

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