To find the maximum height reached by a projectile fired from the surface at an angle above the horizontal, we can follow these steps:
1. Identify the initial speed and angle of projection:
- The initial speed ([tex]\( v_0 \)[/tex]) is [tex]\( 200 \, \text{m/s} \)[/tex].
- The angle of projection ([tex]\( \theta \)[/tex]) is [tex]\( 30^\circ \)[/tex].
2. Break down the initial speed into vertical and horizontal components:
- The vertical component of the initial speed ([tex]\( v_{0y} \)[/tex]) can be calculated using:
[tex]\[
v_{0y} = v_0 \sin(\theta)
\][/tex]
3. Determine the vertical component of the initial speed:
- Using the given values:
[tex]\[
v_{0y} = 200 \times \sin(30^\circ)
\][/tex]
- Since [tex]\(\sin(30^\circ) = 0.5\)[/tex]:
[tex]\[
v_{0y} = 200 \times 0.5 = 100 \, \text{m/s}
\][/tex]
4. Calculate the maximum height reached by the projectile:
- The formula for maximum height ([tex]\( h_{max} \)[/tex]) is derived from the kinematic equation for vertical motion:
[tex]\[
h_{max} = \frac{{v_{0y}^2}}{2g}
\][/tex]
where [tex]\( g \)[/tex] is the acceleration due to gravity ([tex]\( 9.81 \, \text{m/s}^2 \)[/tex]).
5. Substitute the vertical component and gravitational acceleration into the formula:
[tex]\[
h_{max} = \frac{{(100 \, \text{m/s})^2}}{2 \times 9.81 \, \text{m/s}^2}
\][/tex]
6. Calculate the maximum height:
[tex]\[
h_{max} = \frac{{10000}}{19.62}
\][/tex]
[tex]\[
h_{max} \approx 509.68 \, \text{m}
\][/tex]
Therefore, the vertical component of the initial speed is [tex]\( 100 \, \text{m/s} \)[/tex], and the maximum height reached by the projectile is approximately [tex]\( 509.68 \, \text{m} \)[/tex].
