**– Maximum height achieved by the rocket****– For how long did the rocket remain in the air?**

The aim of this question revolves around the understanding and key concepts of **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:

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

## Expert Answer

**Part (a) ****– Maximum height **achieved by the rocket can be calculated **using the following formula**:

\[ h_{ max } \ = \ h_1 \ + \ h_2 \]

Where:

\[ h_1 \ = \ \text{ vertical distance covered during the normal straight line motion } \]

\[ h_2 \ = \ \text{ vertical distance covered during the projectile motion } \]

**Total distance covered** by the rocket **during straight line motion** can be calculated using:

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

\[ S \ = \ ( 200 ) ( 2 ) + \dfrac{ 1 }{ 2 } ( 20 ) ( 2 )^2 \]

\[ S \ = \ 440 \]

**Vertical distance covered** **during straight line motion** can be calculated using the following formula:

\[ h_1 \ = \ S sin \theta \]

\[ h_1 \ = \ ( 440 ) sin( 53^{ \circ } ) \]

\[ h_1 \ = \ 351.40 \]

The** speed at the end** of this part of motion is given by:

\[ v_f \ = \ v_i \ + \ a t \]

\[ v_f \ = \ ( 200 ) \ + \ ( 2 ) ( 2 ) \]

\[ v_f \ = \ 204 \]

**Vertical distance covered during the projectile motion** can be calculated using the following formula:

\[ h_2 \ = \ \dfrac{ v_i^2 \ sin^2 \theta }{ 2 g } \]

Where $ v_i $ is actually the $ v_f $ of the previous part of motion, so:

\[ h_2 \ = \ \dfrac{ ( 204 )^2 \ sin^2 ( 53^{ \circ } ) }{ 2 ( 9.8 ) } \]

\[ \Rightarrow h_2 \ = \ 1354.26 \]

So the **maximum height** will be:

\[ h_{ max } \ = \ h_1 \ + \ h_2 \]

\[ h_{ max } \ = \ 351.40 \ + \ 1354.26 \]

\[ h_{ max } \ = \ 1705.66 \ m \]

**Part (b) – Total flight time **of the rocket can be calculated using the following formula:

\[ t_{ max } \ = \ t_1 \ + \ t_2 \]

Where:

\[ t_1 \ = \ \text{ time taken during the normal straight line motion } \ = \ 2 \ s \]

\[ t_2 \ = \ \text{ time taken covered during the projectile motion } \]

**Time taken during the projectile motion** can be calculated using the following formula:

\[ t_2 \ = \ \dfrac{ 2 v_i \ sin \theta }{ g } \]

\[ t_2 \ = \ \dfrac{ 2 ( 204 ) \ sin ( 53^{ \circ } ) }{ 9.8 } \]

\[ t_2 \ = \ 33.25 \ s \]

So:

\[ t_{ max } \ = \ t_1 \ + \ t_2 \]

\[ t_{ max } \ = \ 2 \ + \ 33.25 \]

\[ t_{ max } \ = \ 35.25 \ s \]

## Numerical Result

\[ h_{ max } \ = \ 1705.66 \ m \]

\[ t_{ max } \ = \ 35.25 \ s \]

## Example

In the same question given above, **How much horizontal distance did the rocket cover during its flight?**

**Maximum horizontal distance **can be calculated using the following formula:

\[ d_{ max } \ = \ d_1 \ + \ d_2 \]

Where:

\[ d_1 \ = \ \text{ horizontal distance covered during the normal straight line motion } \]

\[ d_2 \ = \ \text{ horizontal distance covered during the projectile motion } \]

Total **distance covered** by the rocket **during straight line motion** has already been computed in **part (a) of the above question**:

\[ S \ = \ 440 \]

**Horizontal distance** covered **during the normal straight line motion** can be calculated using the following formula:

\[ d_1 \ = \ S cos \theta \]

\[ d_1 \ = \ ( 440 ) cos( 53^{ \circ } ) \]

\[ d_1 \ = \ 264.80 \]

**Horizontal distance covered during the projectile motion** can be calculated using the following formula:

\[ d_2 \ = \ \dfrac{ v_i^2 \ sin ( 2 \theta ) }{ g } \]

\[ d_2 \ = \ \dfrac{ ( 204 )^2 \ sin ( 2 ( 53^{ \circ } ) ) }{ 9.8 } \]

\[ d_2 \ = \ 4082.03 \]

So:

\[ d_{ max } \ = \ d_1 \ + \ d_2 \]

\[ d_{ max } \ = \ 264.80 \ + \ 4082.03 \]

\[ d_{ max } \ = \ 4346.83 \ m \]