Sure! Let's break down the problem step-by-step to determine the time the projectile is in the air.
### Step-by-Step Solution
1. Given Data:
- Mass of the projectile, [tex]\( m = 200 \)[/tex] g = [tex]\( 0.2 \)[/tex] kg (converted to kilograms).
- Maximum vertical distance (height), [tex]\( h = 10 \)[/tex] m.
- Launch angle, [tex]\( \theta = 30^\circ \)[/tex].
2. Convert the launch angle to radians:
[tex]\[
\theta_{rad} = \theta \times \frac{\pi}{180} = 30 \times \frac{\pi}{180} = 0.5236 \, \text{radians}
\][/tex]
3. Vertical Component of the Initial Velocity:
We use the kinematic equation for vertical motion to find the initial velocity component in the vertical direction, [tex]\( v_{y0} \)[/tex]. The formula for maximum height is:
[tex]\[
h = \frac{v_{y0}^2}{2g}
\][/tex]
where [tex]\( g = 9.81 \, \text{m/s}^2 \)[/tex]. Rearranging for [tex]\( v_{y0} \)[/tex] gives:
[tex]\[
v_{y0} = \sqrt{2gh} = \sqrt{2 \times 9.81 \times 10} = 14.007 \, \text{m/s}
\][/tex]
4. Initial Velocity ( [tex]\( v_{0} \)[/tex] ):
Knowing the vertical component [tex]\( v_{y0} \)[/tex] and the launch angle [tex]\( \theta \)[/tex], we can find the initial velocity [tex]\( v_{0} \)[/tex]:
[tex]\[
v_{y0} = v_{0} \sin(\theta)
\][/tex]
Solving for [tex]\( v_{0} \)[/tex]:
[tex]\[
v_{0} = \frac{v_{y0}}{\sin(\theta)} = \frac{14.007}{\sin(0.5236)} = 28.014 \, \text{m/s}
\][/tex]
5. Total Time of Flight:
The time of flight [tex]\( T \)[/tex] for a projectile is given by:
[tex]\[
T = \frac{2v_{0} \sin(\theta)}{g}
\][/tex]
Substituting the values:
[tex]\[
T = \frac{2 \times 28.014 \times \sin(0.5236)}{9.81} = 2.856 \, \text{seconds}
\][/tex]
Conclusion:
The time the projectile will be in the air is 2.856 seconds.
