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Sagot :
Let's solve the problem step-by-step using the given parameters and the vertical motion equation.
### Step 1: Understand the given data
- Initial velocity ([tex]\( v \)[/tex]): 10.0 meters per second
- Angle of projection ([tex]\( \theta \)[/tex]): [tex]\( 30.0^{\circ} \)[/tex]
- Acceleration due to gravity ([tex]\( g \)[/tex]): 9.81 meters per second squared
- Vertical displacement ([tex]\( \Delta y \)[/tex]): 0 meters (since the ball returns to the same level from which it was kicked)
### Step 2: Convert angle to radians
To use the trigonometric functions in our calculations, convert the angle from degrees to radians:
[tex]\[ \theta_{\text{rad}} = \theta \times \frac{\pi}{180} \][/tex]
Thus,
[tex]\[ \theta_{\text{rad}} = 30.0^{\circ} \times \frac{\pi}{180} = \frac{\pi}{6} \text{ radians} \][/tex]
### Step 3: Calculate the vertical component of the initial velocity
Using the sine function, we find the vertical component of the initial velocity ([tex]\( v_y \)[/tex]):
[tex]\[ v_y = v \sin(\theta_{\text{rad}}) \][/tex]
Substitute in the known values:
[tex]\[ v_y = 10.0 \, \text{m/s} \times \sin\left(\frac{\pi}{6}\right) = 10.0 \, \text{m/s} \times 0.5 = 5.0 \, \text{m/s} \][/tex]
### Step 4: Use the vertical motion equation to determine the time in the air
The vertical motion equation is given by:
[tex]\[ \Delta y = v_y \Delta t + \frac{1}{2} a (\Delta t)^2 \][/tex]
Substitute [tex]\(\Delta y = 0\)[/tex], [tex]\(a = -g\)[/tex], and [tex]\(v_y\)[/tex]:
[tex]\[ 0 = 5.0 \, \text{m/s} \times \Delta t - \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (\Delta t)^2 \][/tex]
Rearranging the equation:
[tex]\[ 0 = 5.0 \, \text{m/s} \times \Delta t - 4.905 \, \text{m/s}^2 \times (\Delta t)^2 \][/tex]
Factor out [tex]\(\Delta t\)[/tex]:
[tex]\[ \Delta t (5.0 \, \text{m/s} - 4.905 \, \text{m/s}^2 \times \Delta t) = 0 \][/tex]
This gives two solutions:
1. [tex]\(\Delta t = 0\)[/tex] (The initial time)
2. [tex]\(5.0 \, \text{m/s} = 4.905 \, \text{m/s}^2 \times \Delta t\)[/tex]
Solving for [tex]\(\Delta t\)[/tex]:
[tex]\[ \Delta t = \frac{5.0 \, \text{m/s}}{4.905 \, \text{m/s}^2} \approx 1.02 \, \text{seconds} \][/tex]
### Step 5: Round to the nearest hundredth
Hence, the time spent in the air is:
[tex]\[ \boxed{1.02} \, \text{seconds} \][/tex]
### Step 1: Understand the given data
- Initial velocity ([tex]\( v \)[/tex]): 10.0 meters per second
- Angle of projection ([tex]\( \theta \)[/tex]): [tex]\( 30.0^{\circ} \)[/tex]
- Acceleration due to gravity ([tex]\( g \)[/tex]): 9.81 meters per second squared
- Vertical displacement ([tex]\( \Delta y \)[/tex]): 0 meters (since the ball returns to the same level from which it was kicked)
### Step 2: Convert angle to radians
To use the trigonometric functions in our calculations, convert the angle from degrees to radians:
[tex]\[ \theta_{\text{rad}} = \theta \times \frac{\pi}{180} \][/tex]
Thus,
[tex]\[ \theta_{\text{rad}} = 30.0^{\circ} \times \frac{\pi}{180} = \frac{\pi}{6} \text{ radians} \][/tex]
### Step 3: Calculate the vertical component of the initial velocity
Using the sine function, we find the vertical component of the initial velocity ([tex]\( v_y \)[/tex]):
[tex]\[ v_y = v \sin(\theta_{\text{rad}}) \][/tex]
Substitute in the known values:
[tex]\[ v_y = 10.0 \, \text{m/s} \times \sin\left(\frac{\pi}{6}\right) = 10.0 \, \text{m/s} \times 0.5 = 5.0 \, \text{m/s} \][/tex]
### Step 4: Use the vertical motion equation to determine the time in the air
The vertical motion equation is given by:
[tex]\[ \Delta y = v_y \Delta t + \frac{1}{2} a (\Delta t)^2 \][/tex]
Substitute [tex]\(\Delta y = 0\)[/tex], [tex]\(a = -g\)[/tex], and [tex]\(v_y\)[/tex]:
[tex]\[ 0 = 5.0 \, \text{m/s} \times \Delta t - \frac{1}{2} \times 9.81 \, \text{m/s}^2 \times (\Delta t)^2 \][/tex]
Rearranging the equation:
[tex]\[ 0 = 5.0 \, \text{m/s} \times \Delta t - 4.905 \, \text{m/s}^2 \times (\Delta t)^2 \][/tex]
Factor out [tex]\(\Delta t\)[/tex]:
[tex]\[ \Delta t (5.0 \, \text{m/s} - 4.905 \, \text{m/s}^2 \times \Delta t) = 0 \][/tex]
This gives two solutions:
1. [tex]\(\Delta t = 0\)[/tex] (The initial time)
2. [tex]\(5.0 \, \text{m/s} = 4.905 \, \text{m/s}^2 \times \Delta t\)[/tex]
Solving for [tex]\(\Delta t\)[/tex]:
[tex]\[ \Delta t = \frac{5.0 \, \text{m/s}}{4.905 \, \text{m/s}^2} \approx 1.02 \, \text{seconds} \][/tex]
### Step 5: Round to the nearest hundredth
Hence, the time spent in the air is:
[tex]\[ \boxed{1.02} \, \text{seconds} \][/tex]
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