Answer :
To solve for the velocity of projection when a stone is projected to achieve a horizontal range of 24 meters and a peak height of 6 meters, we can use principles of projectile motion. Here is the detailed, step-by-step solution for finding the initial velocity [tex]\( U \)[/tex]:
### Step-by-Step Solution:
1. Given Data:
- Horizontal Range ([tex]\( R \)[/tex]) = 24 meters
- Maximum Height ([tex]\( H \)[/tex]) = 6 meters
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.81 m/s²
2. Projectile Motion Equations:
For the vertical motion:
[tex]\[ H = \frac{U^2 \sin^2(\theta)}{2g} \][/tex]
For the horizontal motion:
[tex]\[ R = \frac{U^2 \sin(2\theta)}{g} \][/tex]
3. Equation for Maximum Height:
Rearrange the height formula to solve for [tex]\( U \sin(\theta) \)[/tex]:
[tex]\[ H = \frac{U^2 \sin^2(\theta)}{2g} \][/tex]
[tex]\[ U^2 \sin^2(\theta) = 2gH \][/tex]
[tex]\[ U \sin(\theta) = \sqrt{2gH} \][/tex]
4. Equation for Horizontal Range:
Rearrange the range formula to solve for [tex]\( U \cos(\theta) \)[/tex]:
[tex]\[ R = \frac{U^2 \sin(2\theta)}{g} \][/tex]
Knowing that [tex]\(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\)[/tex]:
[tex]\[ R = \frac{2U^2 \sin(\theta) \cos(\theta)}{g} \][/tex]
[tex]\[ U^2 = \frac{Rg}{2 \sin(\theta) \cos(\theta)} \][/tex]
Since we already calculated [tex]\( U \sin(\theta) \)[/tex]:
[tex]\[ U \cos(\theta) = \frac{R g}{2 \sqrt{2gH}} \][/tex]
5. Calculate Individual Components:
- Calculate [tex]\( U \sin(\theta) \)[/tex]:
[tex]\[ U \sin(\theta) = \sqrt{2 \cdot 9.81 \cdot 6} = \sqrt{117.72} \approx 10.8 \, \text{m/s} \][/tex]
- Calculate [tex]\( U \cos(\theta) \)[/tex]:
[tex]\[ U \cos(\theta) = \frac{24 \cdot 9.81}{2 \cdot \sqrt{2 \cdot 9.81 \cdot 6}} \][/tex]
Substitute the value calculated:
[tex]\[ U \cos(\theta) = \frac{24 \cdot 9.81}{2 \cdot 10.8} \approx 10.9 \, \text{m/s} \][/tex]
6. Resultant Initial Velocity [tex]\( U \)[/tex]:
[tex]\[ U = \sqrt{(U \sin(\theta))^2 + (U \cos(\theta))^2 } \][/tex]
Plug in the values:
[tex]\[ U = \sqrt{10.8^2 + 10.9^2} \][/tex]
[tex]\[ U = \sqrt{116.64 + 118.81} \][/tex]
[tex]\[ U = \sqrt{235.45} \][/tex]
[tex]\[ U \approx 15.3 \, \text{m/s} \][/tex]
Thus, the velocity of projection, [tex]\( U \)[/tex], is approximately [tex]\( 24.3 \, \text{m/s} \)[/tex].
### Step-by-Step Solution:
1. Given Data:
- Horizontal Range ([tex]\( R \)[/tex]) = 24 meters
- Maximum Height ([tex]\( H \)[/tex]) = 6 meters
- Acceleration due to gravity ([tex]\( g \)[/tex]) = 9.81 m/s²
2. Projectile Motion Equations:
For the vertical motion:
[tex]\[ H = \frac{U^2 \sin^2(\theta)}{2g} \][/tex]
For the horizontal motion:
[tex]\[ R = \frac{U^2 \sin(2\theta)}{g} \][/tex]
3. Equation for Maximum Height:
Rearrange the height formula to solve for [tex]\( U \sin(\theta) \)[/tex]:
[tex]\[ H = \frac{U^2 \sin^2(\theta)}{2g} \][/tex]
[tex]\[ U^2 \sin^2(\theta) = 2gH \][/tex]
[tex]\[ U \sin(\theta) = \sqrt{2gH} \][/tex]
4. Equation for Horizontal Range:
Rearrange the range formula to solve for [tex]\( U \cos(\theta) \)[/tex]:
[tex]\[ R = \frac{U^2 \sin(2\theta)}{g} \][/tex]
Knowing that [tex]\(\sin(2\theta) = 2 \sin(\theta) \cos(\theta)\)[/tex]:
[tex]\[ R = \frac{2U^2 \sin(\theta) \cos(\theta)}{g} \][/tex]
[tex]\[ U^2 = \frac{Rg}{2 \sin(\theta) \cos(\theta)} \][/tex]
Since we already calculated [tex]\( U \sin(\theta) \)[/tex]:
[tex]\[ U \cos(\theta) = \frac{R g}{2 \sqrt{2gH}} \][/tex]
5. Calculate Individual Components:
- Calculate [tex]\( U \sin(\theta) \)[/tex]:
[tex]\[ U \sin(\theta) = \sqrt{2 \cdot 9.81 \cdot 6} = \sqrt{117.72} \approx 10.8 \, \text{m/s} \][/tex]
- Calculate [tex]\( U \cos(\theta) \)[/tex]:
[tex]\[ U \cos(\theta) = \frac{24 \cdot 9.81}{2 \cdot \sqrt{2 \cdot 9.81 \cdot 6}} \][/tex]
Substitute the value calculated:
[tex]\[ U \cos(\theta) = \frac{24 \cdot 9.81}{2 \cdot 10.8} \approx 10.9 \, \text{m/s} \][/tex]
6. Resultant Initial Velocity [tex]\( U \)[/tex]:
[tex]\[ U = \sqrt{(U \sin(\theta))^2 + (U \cos(\theta))^2 } \][/tex]
Plug in the values:
[tex]\[ U = \sqrt{10.8^2 + 10.9^2} \][/tex]
[tex]\[ U = \sqrt{116.64 + 118.81} \][/tex]
[tex]\[ U = \sqrt{235.45} \][/tex]
[tex]\[ U \approx 15.3 \, \text{m/s} \][/tex]
Thus, the velocity of projection, [tex]\( U \)[/tex], is approximately [tex]\( 24.3 \, \text{m/s} \)[/tex].
