Question**: A projectile is launched at an angle of 45 degrees with an initial speed of 20 m/s. What is the maximum height reached by the projectile? (Use \( g = 9.8 \, \textm/s^2 \)) - Aurero
Question: A projectile is launched at an angle of 45ð with an initial speed of 20 m/s. What is the maximum height reached by the projectile? (Use ( g = 9.8 , \ ext{m/s}^2 ))
Question: A projectile is launched at an angle of 45ð with an initial speed of 20 m/s. What is the maximum height reached by the projectile? (Use ( g = 9.8 , \ ext{m/s}^2 ))
Understanding Projectile Motion and Maximum Height
Understanding the Context
When a projectile is launched at an angle, its motion can be broken down into horizontal and vertical components. The maximum height is determined solely by the vertical motion, specifically the component of velocity perpendicular to the ground.
Given:
- Launch angle ( \ heta = 45^\circ )
- Initial speed ( v_0 = 20 , \ ext{m/s} )
- Acceleration due to gravity ( g = 9.8 , \ ext{m/s}^2 )
Step 1: Vertical Component of Initial Velocity
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Key Insights
Only the vertical component contributes to reaching maximum height. It is calculated using:
[
v_{y} = v_0 \sin \ heta
]
Substituting ( \ heta = 45^\circ ) and ( \sin 45^\circ = rac{\sqrt{2}}{2} ):
[
v_y = 20 \ imes rac{\sqrt{2}}{2} = 10\sqrt{2} pprox 14.14 , \ ext{m/s}
]
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Step 2: Use Kinematic Equation to Find Maximum Height
At maximum height, the vertical velocity becomes zero (( v_y = 0 )). Using the velocity equation:
[
v_y^2 = u_y^2 - 2gh_{\ ext{max}}
]
Where:
- ( v_y = 0 ) (at peak)
- ( u_y = 10\sqrt{2} , \ ext{m/s} )
- ( h_{\ ext{max}} ) is the maximum height
Rearranging:
[
0 = (10\sqrt{2})^2 - 2 \cdot 9.8 \cdot h_{\ ext{max}}
]
[
0 = 200 - 19.6 \cdot h_{\ ext{max}}
]
Solving for ( h_{\ ext{max}} ):
[
h_{\ ext{max}} = rac{200}{19.6} pprox 10.2 , \ ext{meters}
]
