**Projectile Launch Speed Calculation****Problem:**The highest barrier that a projectile can clear is 23.1 m, when the projectile is launched at an angle of 12° above the horizontal. What is the projectile launch speed?_Number: _______ units_**Explanation:**We have a given angle of launch and the maximum height that the projectile can reach. To find the launch speed, we can use the following kinematic equations and concepts from projectile motion. The highest point in the projectile's trajectory corresponds to the maximum vertical displacement.Key ideas:- \( \theta = 12^\circ \) (launch angle)- \( h_{max} = 23.1 \) m (maximum height)The vertical component of the launch speed \( v_y \) can be found using the equation for vertical motion:\[ h_{max} = \frac{v_{y}^2}{2g} \]where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.Solving for \( v_y \):\[ v_y = \sqrt{2gh_{max}} \]\[ v_y = \sqrt{2 \times 9.8 \times 23.1} \]\[ v_y \approx \sqrt{452.76} \]\[ v_y \approx 21.27 \, \text{m/s} \]The vertical component \( v_y \) is related to the initial launch speed \( v_0 \) by the equation:\[ v_y = v_0 \sin(\theta) \]Therefore,\[ v_0 = \frac{v_y}{\sin(\theta)} \]Substituting the known values:\[ v_0 = \frac{21.27}{\sin(12^\circ)} \]Finally, calculate \( v_0 \):\[ v_0 \approx \frac{21.27}{0.2079} \]\[ v_0 \approx 102.3 \, \text{m/s} \]So, the launch speed \( v_0 \approx 102.3 \, \text{m/s} \).

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Description: **Projectile Launch Speed Calculation****Problem:**The highest barrier that a projectile can clear is 23.1 m, when the projectile is launched at an angle of 12° above the horizontal. What is the projectile launch speed?_Number: _______ units_**Expla

Given data: Maximum height (Hmax) = 23.1 m Projectile angle (θ) = 12°, above horizontal Projectile…

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The highest barrier that a progechle can clear is 23.1m, When the projectile is launched at an cungle of i2 above horizuntal, what is the prosectile launch speed ne number units

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**Projectile Launch Speed Calculation**

**Problem:**
The highest barrier that a projectile can clear is 23.1 m, when the projectile is launched at an angle of 12° above the horizontal. What is the projectile launch speed?

_Number: _______ units_

**Explanation:**
We have a given angle of launch and the maximum height that the projectile can reach. To find the launch speed, we can use the following kinematic equations and concepts from projectile motion. The highest point in the projectile's trajectory corresponds to the maximum vertical displacement.

Key ideas:
- \( \theta = 12^\circ \) (launch angle)
- \( h_{max} = 23.1 \) m (maximum height)

The vertical component of the launch speed \( v_y \) can be found using the equation for vertical motion:
\[ h_{max} = \frac{v_{y}^2}{2g} \]
where \( g = 9.8 \, \text{m/s}^2 \) is the acceleration due to gravity.

Solving for \( v_y \):
\[ v_y = \sqrt{2gh_{max}} \]
\[ v_y = \sqrt{2 \times 9.8 \times 23.1} \]
\[ v_y \approx \sqrt{452.76} \]
\[ v_y \approx 21.27 \, \text{m/s} \]

The vertical component \( v_y \) is related to the initial launch speed \( v_0 \) by the equation:
\[ v_y = v_0 \sin(\theta) \]

Therefore,
\[ v_0 = \frac{v_y}{\sin(\theta)} \]

Substituting the known values:
\[ v_0 = \frac{21.27}{\sin(12^\circ)} \]

Finally, calculate \( v_0 \):
\[ v_0 \approx \frac{21.27}{0.2079} \]
\[ v_0 \approx 102.3 \, \text{m/s} \]

So, the launch speed \( v_0 \approx 102.3 \, \text{m/s} \).

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