An artillery shell is fired with an initial velocity of 300 m/s at 55 degrees above the horizontal. to clear an avalanche, it explodes on a mountainside 42 sec after firing. what is the x-coordinate of the shell where it explodes, relative to its firing point?

An artillery shell is fired with an initial velocity of 300 m/s at 55 degrees above the horizontal. to clear an avalanche, it explodes on a mountainside 42 sec after firing. What is the y-coordinate of the shell where it explodes, relative to its firing point?

ok, so you have to find the cos of x-axis: v*cos=(300m/s)cos 55 degrees, and then the sin of your y-axis: v*sin= (300m/s)sin 55 degrees. The vx (x-axis)is always constant.

Then you have to use the formula: x=x(i)*t+ 1/2axt^2 to get the x- coordinate, y=y(i)*t +1/2ayt^2. That'll give you the y- coordination. Hope that helps you..Good luck!

To find the x-coordinate of the point where the artillery shell explodes, we need to calculate the horizontal distance traveled by the shell in 42 seconds.

First, we can split the initial velocity into its horizontal and vertical components. The horizontal component can be found using the equation:

Vx = V * cos(θ)

where V is the initial velocity and θ is the angle above the horizontal.

Vx = 300 m/s * cos(55°)
≈ 300 m/s * 0.5736
≈ 172.08 m/s

Now, to find the x-coordinate, we can use the equation:

x = Vx * t

where t is the time traveled.

x = 172.08 m/s * 42 s
≈ 7235.36 m

Therefore, the x-coordinate of the point where the shell explodes, relative to its firing point, is approximately 7235.36 meters.

To find the y-coordinate of the point where the shell explodes, relative to its firing point, we need to calculate the vertical distance traveled by the shell in 42 seconds.

The vertical component of the initial velocity can be found using the equation:

Vy = V * sin(θ)

where V is the initial velocity and θ is the angle above the horizontal.

Vy = 300 m/s * sin(55°)
≈ 300 m/s * 0.8192
≈ 245.76 m/s

Now, we can use kinematic equations to calculate the vertical distance traveled. The equation we can use is:

y = Vy * t - (1/2) * g * t^2

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

y = 245.76 m/s * 42 s - (1/2) * 9.8 m/s^2 * (42 s)^2
≈ 10322.32 m - (1/2) * 9.8 m/s^2 * 1764 s^2
≈ 10322.32 m - 86268 m
≈ -75945.68 m

However, this negative value indicates that the shell has gone below the initial height, which is not possible. Therefore, there might be an issue with the given information or calculations.

Please double-check the given data or question and try again.

To solve for the x-coordinate and the y-coordinate of the shell where it explodes, we need to first break down the initial velocity into its horizontal and vertical components.

The horizontal component (Vx) of the velocity can be found using the formula:
Vx = V * cos(theta)

where V is the initial velocity of 300 m/s and theta is the launch angle of 55 degrees.

Vx = 300 * cos(55)
Vx ≈ 300 * 0.5736
Vx ≈ 172.08 m/s

The vertical component (Vy) of the velocity can be found using the formula:
Vy = V * sin(theta)

Vy = 300 * sin(55)
Vy ≈ 300 * 0.8192
Vy ≈ 245.76 m/s

Now, we can use the horizontal component to find the x-coordinate of the shell where it explodes after 42 seconds. Since there are no horizontal forces acting on the shell, the horizontal velocity remains constant.

x = Vx * t
x = 172.08 * 42
x ≈ 7223.36 m

Therefore, the x-coordinate of the shell where it explodes, relative to its firing point, is approximately 7223.36 meters.

For the y-coordinate, we can use the vertical component to find the displacement of the shell where it explodes.

The formula for vertical displacement (h) is:
h = Vy * t + (1/2) * g * t^2

where g is the acceleration due to gravity, which is approximately 9.8 m/s^2.

h = 245.76 * 42 + (1/2) * 9.8 * (42^2)
h ≈ 10336.32 + (1/2) * 9.8 * 1764
h ≈ 10336.32 + (1/2) * 9.8 * 3094.32
h ≈ 10336.32 + 60224.856
h ≈ 70561.176 m

Therefore, the y-coordinate of the shell where it explodes, relative to its firing point, is approximately 70561.176 meters.