Sure, let's go through the detailed, step-by-step solution for each part of the problem.
### Part i: Calculate the vertical displacement at a time of 2 seconds
To find the vertical displacement of a particle under gravity, we use the kinematic equation for vertical motion under uniform acceleration (gravity, [tex]\( g \)[/tex]), starting from rest vertically. The equation for vertical displacement [tex]\( s \)[/tex] is given by:
[tex]\[ s = \frac{1}{2} g t^2 \][/tex]
where:
- [tex]\( g \)[/tex] is the acceleration due to gravity ([tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex])
- [tex]\( t \)[/tex] is the time in seconds ([tex]\( t = 2 \, \text{s} \)[/tex])
Plugging in the values:
[tex]\[ s = \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (2 \, \text{s})^2 \][/tex]
[tex]\[ s = \frac{1}{2} \times 9.81 \times 4 \][/tex]
[tex]\[ s = 19.62 \, \text{meters} \][/tex]
So, the vertical displacement at a time of 2 seconds is [tex]\( 19.62 \, \text{meters} \)[/tex].
### Part ii: Calculate the time to reach a vertical drop of 5 meters
To find the time it takes for the particle to reach a vertical drop of 5 meters, we'll use the same kinematic equation, solved for time [tex]\( t \)[/tex]:
[tex]\[ s = \frac{1}{2} g t^2 \][/tex]
To solve for [tex]\( t \)[/tex], rearrange the equation:
[tex]\[ t^2 = \frac{2s}{g} \][/tex]
Taking the square root of both sides:
[tex]\[ t = \sqrt{\frac{2s}{g}} \][/tex]
where:
- [tex]\( s = 5 \, \text{meters} \)[/tex]
- [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]
Plugging in the values:
[tex]\[ t = \sqrt{\frac{2 \times 5 \, \text{meters}}{9.81 \, \text{m/s}^2}} \][/tex]
[tex]\[ t = \sqrt{\frac{10}{9.81}} \][/tex]
[tex]\[ t \approx 1.0096 \, \text{seconds} \][/tex]
So, the time to reach a vertical drop of 5 meters is approximately [tex]\( 1.0096 \, \text{seconds} \)[/tex].
### Summary
- The vertical displacement at a time of 2 seconds is [tex]\( 19.62 \, \text{meters} \)[/tex].
- The time to reach a vertical drop of 5 meters is approximately [tex]\( 1.0096 \, \text{seconds} \)[/tex].
