A cannon ball whose barrel is angled at 40˚ from the horizontal fires a shell with a muzzle velocity of 350 m/s. What is the horizontal and vertical component of the muzzle velocity and what is the range of the shell?
v₀⒳=v₀•cosα
v₀⒴=v₀•sinα
L=vₒ²•sin2α/g,
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To find the horizontal and vertical components of the muzzle velocity, we can use trigonometry. The horizontal component represents the velocity in the horizontal direction (along the x-axis), while the vertical component represents the velocity in the vertical direction (along the y-axis).
Given:
Muzzle velocity (v) = 350 m/s
Angle (θ) = 40˚
To find the horizontal component (v_x), we need to use the cosine function:
v_x = v × cos(θ)
Substituting the given values:
v_x = 350 m/s × cos(40˚)
Calculating this value:
v_x ≈ 350 m/s × 0.766 = 268.1 m/s
Therefore, the horizontal component of the muzzle velocity is approximately 268.1 m/s.
To find the vertical component (v_y), we use the sine function:
v_y = v × sin(θ)
Substituting the given values:
v_y = 350 m/s × sin(40˚)
Calculating this value:
v_y ≈ 350 m/s × 0.642 = 225.7 m/s
So, the vertical component of the muzzle velocity is approximately 225.7 m/s.
To calculate the range of the shell, we need to consider the time of flight. The time of flight can be determined using the vertical component of velocity and the acceleration due to gravity (g).
Using the equation:
Δy = v_y × t + (1/2) × g × t^2
Since the cannonball is fired at an angle, the initial vertical position is zero and the final vertical position (Δy) is also zero at the point of landing. We can set up the equation as follows:
0 = v_y × t + (1/2) × g × t^2
Rearranging the equation, we get:
(1/2) × g × t^2 = v_y × t
Simplifying, we have:
(1/2) × g × t = v_y
t = (2 × v_y) / g
Using the acceleration due to gravity (g ≈ 9.8 m/s^2), and the vertical component of the muzzle velocity (v_y ≈ 225.7 m/s):
t ≈ (2 × 225.7 m/s) / 9.8 m/s^2 ≈ 46.2 s
The time of flight is approximately 46.2 seconds.
The range of the shell can be calculated using the horizontal component of velocity and the time of flight:
Range = v_x × t
Substituting the values:
Range ≈ 268.1 m/s × 46.2 s ≈ 12,373.2 m
Therefore, the range of the shell is approximately 12,373.2 meters.