 # A rocket is launched at an angle of 53 degrees above the horizontal with an initial speed of 200 m/s. The rocket moves for 2.00 s along its initial line of motion with an acceleration of 20.0 m/s^2. At this time, its engines fail and the rocket proceeds to move as a projectile. Calculate the following quantities. – 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:

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

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:

$d_{ max } \ = \ d_1 \ + \ d_2$

$d_{ max } \ = \ 264.80 \ + \ 4082.03$

$d_{ max } \ = \ 4346.83 \ m$