Answer :
To determine how far the rock dropped, we need to follow a series of steps involving the concepts of projectile motion. In particular, we will use the rock’s horizontal velocity and the horizontal distance it traveled to find the time of flight, and then use that time to calculate the vertical distance it fell under the influence of gravity.
1. Find the time of flight:
Given:
- Horizontal velocity [tex]\( v_x = 2.87 \, \text{m/s} \)[/tex]
- Horizontal distance [tex]\( d_x = 5.32 \, \text{m} \)[/tex]
The time [tex]\( t \)[/tex] it takes for the rock to travel the horizontal distance can be found using the formula:
[tex]\[ t = \frac{d_x}{v_x} \][/tex]
Substituting the given values:
[tex]\[ t = \frac{5.32 \, \text{m}}{2.87 \, \text{m/s}} \approx 1.8537 \, \text{s} \][/tex]
2. Calculate the vertical distance:
To find the vertical distance [tex]\( \Delta y \)[/tex] the rock fell, we use the kinematic equation for vertical motion under uniform acceleration (gravity):
[tex]\[ \Delta y = \frac{1}{2} g t^2 \][/tex]
where [tex]\( g \)[/tex] is the acceleration due to gravity. We assume [tex]\( g \approx 9.81 \, \text{m/s}^2 \)[/tex].
Using the time [tex]\( t \)[/tex] we calculated:
[tex]\[ \Delta y = \frac{1}{2} \cdot 9.81 \, \text{m/s}^2 \cdot (1.8537 \, \text{s})^2 \][/tex]
[tex]\[ \Delta y \approx \frac{1}{2} \cdot 9.81 \cdot 3.4362 \approx 16.8538 \, \text{m} \][/tex]
Therefore, the vertical distance [tex]\( \Delta y \)[/tex] that the rock dropped is approximately 16.8538 meters.
1. Find the time of flight:
Given:
- Horizontal velocity [tex]\( v_x = 2.87 \, \text{m/s} \)[/tex]
- Horizontal distance [tex]\( d_x = 5.32 \, \text{m} \)[/tex]
The time [tex]\( t \)[/tex] it takes for the rock to travel the horizontal distance can be found using the formula:
[tex]\[ t = \frac{d_x}{v_x} \][/tex]
Substituting the given values:
[tex]\[ t = \frac{5.32 \, \text{m}}{2.87 \, \text{m/s}} \approx 1.8537 \, \text{s} \][/tex]
2. Calculate the vertical distance:
To find the vertical distance [tex]\( \Delta y \)[/tex] the rock fell, we use the kinematic equation for vertical motion under uniform acceleration (gravity):
[tex]\[ \Delta y = \frac{1}{2} g t^2 \][/tex]
where [tex]\( g \)[/tex] is the acceleration due to gravity. We assume [tex]\( g \approx 9.81 \, \text{m/s}^2 \)[/tex].
Using the time [tex]\( t \)[/tex] we calculated:
[tex]\[ \Delta y = \frac{1}{2} \cdot 9.81 \, \text{m/s}^2 \cdot (1.8537 \, \text{s})^2 \][/tex]
[tex]\[ \Delta y \approx \frac{1}{2} \cdot 9.81 \cdot 3.4362 \approx 16.8538 \, \text{m} \][/tex]
Therefore, the vertical distance [tex]\( \Delta y \)[/tex] that the rock dropped is approximately 16.8538 meters.
