Answer :
Certainly! Let's calculate the time spent in the air for Norma's soccer ball step-by-step.
### Step 1: Identify the Given Information
1. Initial velocity ([tex]\( v \)[/tex]): 10.0 meters per second
2. Angle ([tex]\( \theta \)[/tex]): 30 degrees
3. Gravitational acceleration ([tex]\( g \)[/tex]): 9.81 meters per second squared
4. Vertical displacement ([tex]\( \Delta y \)[/tex]): 0 meters (since the ball returns to ground level)
### Step 2: Convert the Angle to Radians
To use trigonometric functions, we need to convert the angle from degrees to radians.
[tex]\[ \theta_{\text{rad}} = \theta \times \frac{\pi}{180} \][/tex]
[tex]\[ \theta_{\text{rad}} = 30 \times \frac{\pi}{180} = \frac{\pi}{6} \][/tex]
### Step 3: Calculate the Vertical Component of the Initial Velocity
The initial velocity in the vertical direction ([tex]\( v_y \)[/tex]) can be calculated using the sine function:
[tex]\[ v_y = v \sin(\theta_{\text{rad}}) \][/tex]
[tex]\[ v_y = 10 \sin\left(\frac{\pi}{6}\right) \][/tex]
Using the fact that [tex]\(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\)[/tex]:
[tex]\[ v_y = 10 \times \frac{1}{2} = 5 \text{ m/s} \][/tex]
### Step 4: Use the Vertical Motion Equation
The vertical displacement ([tex]\( \Delta y \)[/tex]) equation is given by:
[tex]\[ \Delta y = v_y \Delta t + \frac{1}{2} a \Delta t^2 \][/tex]
Since [tex]\(\Delta y\)[/tex] is 0, we set up the quadratic equation:
[tex]\[ 0 = 5\Delta t + \frac{1}{2}(-9.81)(\Delta t^2) \][/tex]
[tex]\[ 0 = 5\Delta t - 4.905(\Delta t^2) \][/tex]
### Step 5: Simplify and Solve the Quadratic Equation
Rewriting the equation in a standard quadratic form:
[tex]\[ 4.905(\Delta t^2) - 5\Delta t = 0 \][/tex]
Factor out [tex]\(\Delta t\)[/tex]:
[tex]\[ \Delta t(4.905\Delta t - 5) = 0 \][/tex]
This gives us two solutions:
[tex]\[ \Delta t = 0 \text{ seconds} \][/tex]
[tex]\[ 4.905\Delta t = 5 \][/tex]
[tex]\[ \Delta t = \frac{5}{4.905} \approx 1.019 \text{ seconds} \][/tex]
Given that [tex]\(\Delta t = 0\)[/tex] represents the initial kick, we disregard it and take the positive time value.
### Step 6: Round the Time to the Nearest Hundredth
The time spent in the air:
[tex]\[ \Delta t \approx 1.019 \text{ seconds} \][/tex]
Rounding to the nearest hundredth:
[tex]\[ \Delta t \approx 1.02 \text{ seconds} \][/tex]
Therefore, the time spent in the air is approximately 1.02 seconds.
### Step 1: Identify the Given Information
1. Initial velocity ([tex]\( v \)[/tex]): 10.0 meters per second
2. Angle ([tex]\( \theta \)[/tex]): 30 degrees
3. Gravitational acceleration ([tex]\( g \)[/tex]): 9.81 meters per second squared
4. Vertical displacement ([tex]\( \Delta y \)[/tex]): 0 meters (since the ball returns to ground level)
### Step 2: Convert the Angle to Radians
To use trigonometric functions, we need to convert the angle from degrees to radians.
[tex]\[ \theta_{\text{rad}} = \theta \times \frac{\pi}{180} \][/tex]
[tex]\[ \theta_{\text{rad}} = 30 \times \frac{\pi}{180} = \frac{\pi}{6} \][/tex]
### Step 3: Calculate the Vertical Component of the Initial Velocity
The initial velocity in the vertical direction ([tex]\( v_y \)[/tex]) can be calculated using the sine function:
[tex]\[ v_y = v \sin(\theta_{\text{rad}}) \][/tex]
[tex]\[ v_y = 10 \sin\left(\frac{\pi}{6}\right) \][/tex]
Using the fact that [tex]\(\sin\left(\frac{\pi}{6}\right) = \frac{1}{2}\)[/tex]:
[tex]\[ v_y = 10 \times \frac{1}{2} = 5 \text{ m/s} \][/tex]
### Step 4: Use the Vertical Motion Equation
The vertical displacement ([tex]\( \Delta y \)[/tex]) equation is given by:
[tex]\[ \Delta y = v_y \Delta t + \frac{1}{2} a \Delta t^2 \][/tex]
Since [tex]\(\Delta y\)[/tex] is 0, we set up the quadratic equation:
[tex]\[ 0 = 5\Delta t + \frac{1}{2}(-9.81)(\Delta t^2) \][/tex]
[tex]\[ 0 = 5\Delta t - 4.905(\Delta t^2) \][/tex]
### Step 5: Simplify and Solve the Quadratic Equation
Rewriting the equation in a standard quadratic form:
[tex]\[ 4.905(\Delta t^2) - 5\Delta t = 0 \][/tex]
Factor out [tex]\(\Delta t\)[/tex]:
[tex]\[ \Delta t(4.905\Delta t - 5) = 0 \][/tex]
This gives us two solutions:
[tex]\[ \Delta t = 0 \text{ seconds} \][/tex]
[tex]\[ 4.905\Delta t = 5 \][/tex]
[tex]\[ \Delta t = \frac{5}{4.905} \approx 1.019 \text{ seconds} \][/tex]
Given that [tex]\(\Delta t = 0\)[/tex] represents the initial kick, we disregard it and take the positive time value.
### Step 6: Round the Time to the Nearest Hundredth
The time spent in the air:
[tex]\[ \Delta t \approx 1.019 \text{ seconds} \][/tex]
Rounding to the nearest hundredth:
[tex]\[ \Delta t \approx 1.02 \text{ seconds} \][/tex]
Therefore, the time spent in the air is approximately 1.02 seconds.
