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