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Sagot :
Sure, let's break this down step-by-step to understand how we find the average velocity of a projectile between the instants it crosses half its maximum height.
### Step 1: Given Information
1. Initial Speed (u): 20 m/s
2. Angle of Projection (θ): 45 degrees
3. Acceleration due to Gravity (g): 9.8 m/s²
### Step 2: Maximum Height Calculation (H)
The maximum height [tex]\( H \)[/tex] of a projectile is given by the formula:
[tex]\[ H = \frac{{u^2 \sin^2 \theta}}{{2g}} \][/tex]
Using the values provided, we calculate:
[tex]\[ H = 10.204081632653057 \text{ m} \][/tex]
### Step 3: Half of the Maximum Height (half_H)
To find half the maximum height:
[tex]\[ \text{half\_H} = \frac{H}{2} = \frac{10.204081632653057}{2} = 5.102040816326529 \text{ m} \][/tex]
### Step 4: Time to Reach Maximum Height
The time to reach the maximum height [tex]\( t_{max} \)[/tex] can be calculated using:
[tex]\[ t_{max} = \frac{{u \sin \theta}}{g} \][/tex]
Using these values, we calculate:
[tex]\[ t_{max} = 1.4430750636460152 \text{ s} \][/tex]
### Step 5: Time to Reach Half Maximum Height (t1 and t2)
To find the time at which the projectile reaches half maximum height, we solve the equation of motion:
[tex]\[ \text{half\_H} = u \sin \theta \cdot t - \frac{1}{2} g t^2 \][/tex]
This quadratic equation has two solutions, [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex]:
[tex]\[ t_1 = 0.4226669003807091 \text{ s} \][/tex]
[tex]\[ t_2 = 2.4634832269113214 \text{ s} \][/tex]
### Step 6: Average Velocity Calculation
Average velocity [tex]\( v_{avg} \)[/tex] between the time instants [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] when the projectile crosses half the maximum height is given by:
The horizontal component of the velocity remains constant, as there's no acceleration in the horizontal direction:
[tex]\[ v_{avg} = u \cos \theta \][/tex]
Using these values, we calculate:
[tex]\[ v_{avg} = 14.142135623730951 \text{ m/s} \][/tex]
### Final Average Velocity
The average velocity of the projectile between the instants it crosses half the maximum height is:
[tex]\[ \boxed{14.142135623730951 \text{ m/s}} \][/tex]
### Step 1: Given Information
1. Initial Speed (u): 20 m/s
2. Angle of Projection (θ): 45 degrees
3. Acceleration due to Gravity (g): 9.8 m/s²
### Step 2: Maximum Height Calculation (H)
The maximum height [tex]\( H \)[/tex] of a projectile is given by the formula:
[tex]\[ H = \frac{{u^2 \sin^2 \theta}}{{2g}} \][/tex]
Using the values provided, we calculate:
[tex]\[ H = 10.204081632653057 \text{ m} \][/tex]
### Step 3: Half of the Maximum Height (half_H)
To find half the maximum height:
[tex]\[ \text{half\_H} = \frac{H}{2} = \frac{10.204081632653057}{2} = 5.102040816326529 \text{ m} \][/tex]
### Step 4: Time to Reach Maximum Height
The time to reach the maximum height [tex]\( t_{max} \)[/tex] can be calculated using:
[tex]\[ t_{max} = \frac{{u \sin \theta}}{g} \][/tex]
Using these values, we calculate:
[tex]\[ t_{max} = 1.4430750636460152 \text{ s} \][/tex]
### Step 5: Time to Reach Half Maximum Height (t1 and t2)
To find the time at which the projectile reaches half maximum height, we solve the equation of motion:
[tex]\[ \text{half\_H} = u \sin \theta \cdot t - \frac{1}{2} g t^2 \][/tex]
This quadratic equation has two solutions, [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex]:
[tex]\[ t_1 = 0.4226669003807091 \text{ s} \][/tex]
[tex]\[ t_2 = 2.4634832269113214 \text{ s} \][/tex]
### Step 6: Average Velocity Calculation
Average velocity [tex]\( v_{avg} \)[/tex] between the time instants [tex]\( t_1 \)[/tex] and [tex]\( t_2 \)[/tex] when the projectile crosses half the maximum height is given by:
The horizontal component of the velocity remains constant, as there's no acceleration in the horizontal direction:
[tex]\[ v_{avg} = u \cos \theta \][/tex]
Using these values, we calculate:
[tex]\[ v_{avg} = 14.142135623730951 \text{ m/s} \][/tex]
### Final Average Velocity
The average velocity of the projectile between the instants it crosses half the maximum height is:
[tex]\[ \boxed{14.142135623730951 \text{ m/s}} \][/tex]
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