An artillery shell is launched on a flat, horizontal field at an angle of α = 31.5° with respect to the horizontal and with an initial speed of v0 = 266 m/s.
What is the horizontal distance covered by the shell after 9.44 s of flight?
What is the height of the shell at this moment?
I think you need a real tutor. Every question you submit, has been responded to with directions, yet you keep coming back saying you don't understand. I have reread some of my responses, putting them in to algebra I terms for you to finish, yet you say you don't understand. At this point, you need a tutor. Discuss this immediately with your parents, as delaying it will just put you in a hole that is difficult to get out of.
Good luck.
To find the horizontal distance covered by the shell after 9.44 s of flight, we can use the formula for horizontal distance:
𝑑 = 𝑣₀ * 𝑡 * cos(𝛼)
where:
𝑑 = horizontal distance covered by the shell
𝑣₀ = initial speed of the shell
𝑡 = time of flight
𝛼 = launch angle
Given:
𝑣₀ = 266 m/s
𝑡 = 9.44 s
𝛼 = 31.5°
Substituting the given values into the formula, we have:
𝑑 = 266 * 9.44 * cos(31.5°)
Now, let's evaluate this expression using a calculator:
𝑑 ≈ 2191.36 m
So, the horizontal distance covered by the shell after 9.44 s of flight is approximately 2191.36 meters.
To find the height of the shell at this moment, we can use the formula for vertical distance or height:
ℎ = 𝑣₀ * 𝑡 * sin(𝛼) - (1/2) * 𝑔 * 𝑡²
where:
ℎ = height of the shell
𝑣₀ = initial speed of the shell
𝑡 = time of flight
𝛼 = launch angle
𝑔 = acceleration due to gravity (approximately 9.8 m/s²)
Given:
𝑣₀ = 266 m/s
𝑡 = 9.44 s
𝛼 = 31.5°
𝑔 = 9.8 m/s²
Substituting the given values into the formula, we have:
ℎ = 266 * 9.44 * sin(31.5°) - (1/2) * 9.8 * (9.44)²
Now, let's evaluate this expression using a calculator:
ℎ ≈ 270.811 m
So, the height of the shell at this moment is approximately 270.811 meters.