**Determine the following quantities:**

**– The horizontal and vertical components of the velocity vector.**

**– The maximum height reached by the projectile above the launch point.**

The **aim of this question** is to understand the different **parameters** during **2D projectile motion.**

The most important parameters during the flight of a projectile are its **range, time 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:

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

The same problem can be solved with the fundamental **equations of motion**. Which are given below:

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

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

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

## Expert Answer

**Given that:**

\[ v_i \ =\ 65 \ m/s \]

\[ \theta \ =\ 37^{ \circ } \]

\[ h_i \ =\ 125 \ m \]

**Part (a) – The horizontal and vertical components of the velocity vector.**

\[ v_{i_{x}} \ =\ v_i cos ( \theta ) \ = \ 65 cos( 37^{ \circ } ) \ = \ 52 \ m/s \]

\[ v_{i_{y}} \ =\ v_i sin ( \theta ) \ = \ 65 sin( 37^{ \circ } ) \ = \ 39 \ m/s \]

**Part (b) – The maximum height reached by the projectile above the launch point.**

For upward motion:

\[ v_i \ =\ 39 \ m/s \]

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

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

Using 3rd equation of motion:

\[ S \ = \ \dfrac{ v_f^2 – v_i^2 }{ 2a } \]

\[ S \ = \ \dfrac{ 0^2 – 39^2 }{ 2(-9.8) } \]

\[ S \ = \ \dfrac{ 1521 }{ 19.6 } \]

\[ S \ = \ 77.60 \ m \]

## Numerical Result

**Part (a) – The horizontal and vertical components of the velocity vector:**

\[ v_{i_{x}} \ = \ 52 \ m/s \]

\[ v_{i_{y}} \ = \ 39 \ m/s \]

**Part (b) – The maximum height reached by the projectile above the launch point:**

\[ S \ = \ 77.60 \ m \]

## Example

For the same projectile given in the question above, find the**Â time elapsed before hitting the ground level.**

**For upward motion:**

\[ v_i \ =\ 39 \ m/s \]

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

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

**Using 1st equation of motion:**

\[ t_1 \ = \ \dfrac{ v_f – v_i }{ a } \]

\[ t_1 \ = \ \dfrac{ 0 – 39 }{ -9.8 } \]

\[ t_1 \ = \ 3.98 \ s \]

**For downward motion:**

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

\[ S \ = \ 77.60 + 125 \ = \ 180.6 \ m \]

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

**Using 2nd equation of motion:**

\[ S \ = \ v_{i} t_2 + \dfrac{ 1 }{ 2 } a t_2^2 \]

\[ 180.6 \ = \ (0) t_2 + \dfrac{ 1 }{ 2 } ( 9.8 ) t_2^2 \]

\[ 180.6 \ = \ \dfrac{ 1 }{ 2 } ( 9.8 ) t_2^2 \]

\[ t_2^2 \ = \ 36.86 \]

\[ t_2 \ = \ 6.07 \ s \]

**So the total time:**

\[ t \ = \ t_1 + t_2 \ = \ 3.98Â + 6.07 \ = \ 10.05 \ s \]