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A cannon fires a shell at 180 m/s at an angle of 25° to the horizontal. How far does it travel on level ground? (Show your work).

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Answer:

The shell travels approximately 1245.48 meters on level ground.

Step-by-step explanation:

To find the horizontal distance the shell travels on level ground, we can use the equations of projectile motion. The horizontal and vertical motion of the shell are independent of each other.

The initial velocity can be broken down into horizontal and vertical components:

Initial velocity (V₀) = 180 m/s

Angle of projection (θ) = 25°

Horizontal component of velocity (V₀x) = V₀ * cos(θ)

Vertical component of velocity (V₀y) = V₀ * sin(θ)

In this case:

V₀x = 180 m/s * cos(25°) ≈ 162.92 m/s

V₀y = 180 m/s * sin(25°) ≈ 75.00 m/s

Now, let's focus on the horizontal motion of the shell. The time of flight (T) for the projectile can be calculated using the vertical motion equation:

T = V₀y / g

where g is the acceleration due to gravity (approximately 9.81 m/s²). Plugging in the values:

T = 75.00 m/s / 9.81 m/s² ≈ 7.64 seconds

Now, we can find the horizontal distance traveled (range) using the horizontal motion equation:

Range = V₀x * T

Range = 162.92 m/s * 7.64 s ≈ 1245.48 meters

So, the shell travels approximately 1245.48 meters on level ground.

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User Esben Eickhardt
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