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?

To find the x-coordinate and y-coordinate of the shell where it explodes, we can break down the motion into horizontal and vertical components.

Given:
Initial velocity (v₀) = 300 m/s
Launch angle (θ) = 55 degrees
Time of flight (t) = 42 seconds

1. Finding the x-coordinate:
The x-coordinate of the shell's explosion point can be found by considering the horizontal component of the motion. The horizontal velocity (vₓ) remains constant throughout the flight because there are no horizontal forces acting on the shell.

The formula to calculate the horizontal displacement (x) is:
x = vₓ * t

To find the horizontal velocity (vₓ), we need to determine the horizontal component of the initial velocity (v₀) using trigonometry:
vₓ = v₀ * cos(θ)

Then, we substitute the values into the equation to find the x-coordinate:
x = (v₀ * cos(θ)) * t

2. Finding the y-coordinate:
The y-coordinate of the shell's explosion point can be found by considering the vertical component of the motion. The vertical velocity (vᵧ) changes due to the acceleration from gravity.

The formula to calculate the vertical displacement (y) is:
y = vᵧ₀ * t + (1/2) * g * t²

To find the vertical velocity at the start (vᵧ₀), we need to determine the vertical component of the initial velocity (v₀) using trigonometry:
vᵧ₀ = v₀ * sin(θ)

Then, we substitute the values into the equation to find the y-coordinate:
y = (v₀ * sin(θ)) * t + (1/2) * g * t²

Note: In this case, we don't have the value of acceleration due to gravity (g). Assuming standard gravity, we can use g = 9.8 m/s².

Now, let's calculate the x-coordinate and y-coordinate using the given values:

x = (300 * cos(55)) * 42
y = (300 * sin(55)) * 42 + (1/2) * 9.8 * (42)²

Evaluating these calculations will give us the answer for both the x-coordinate and y-coordinate of the shell's explosion point relative to its firing point.