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Answer :
a) we can use the horizontal distance and time of flight to find the initial speed:
v₀ = Δx / t = 1330 / 4.04 ≈ 329.70 m/s
b) The magnitude of the displacement is also approximately 1330 m.
a. To determine the initial speed of the shell, we can use the kinematic equation:
Δx = v₀t + (1/2)at²
Where Δx is the horizontal distance traveled (1330 m), v₀ is the initial velocity, t is the time of flight, and a is the acceleration due to gravity (-9.8 m/s²). We can ignore the vertical motion since it doesn't affect the horizontal distance traveled.
Since the shell is fired horizontally, its initial vertical velocity is zero. Therefore, the time of flight can be determined from the vertical motion:
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Δy = v₀y*t + (1/2)gt²
Where Δy is the vertical distance traveled (80 m), v₀y is the initial vertical velocity (zero), and g is the acceleration due to gravity (-9.8 m/s²). Solving for t:
t = sqrt((2Δy)/g) = sqrt((2*80)/9.8) ≈ 4.04 s
Now we can use the horizontal distance and time of flight to find the initial speed:
v₀ = Δx / t = 1330 / 4.04 ≈ 329.70 m/s
b. The horizontal component of the final velocity is the same as the initial velocity, since there is no horizontal acceleration. Therefore, the speed of the shell as it hits the ground is also approximately 329.70 m/s.
c. The angle between the final velocity and the horizontal can be found using trigonometry. We can use the vertical distance traveled and the time of flight to find the final vertical velocity:
vfy = gt = 9.8 * 4.04 ≈ 39.59 m/s
The magnitude of the final velocity is the square root of the sum of the squares of the horizontal and vertical components:
vf = sqrt(v₀² + vfy²) ≈ 340.51 m/s
The angle between the final velocity and the horizontal can be found using the inverse tangent function:
θ = arctan(vfy / v₀) ≈ 0.120 radians ≈ 6.87 degrees
d. The displacement of the shell is the vector difference between its initial and final positions. Since the shell starts and ends at the same height, we only need to consider the horizontal displacement:
Δx = v₀t = 329.70 * 4.04 ≈ 1330 m
So the magnitude of the displacement is also approximately 1330 m.
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