An artillery shell is fired with an initial velocity of 300 m/s at 65.0° above the horizontal. To clear an avalanche, it explodes on a mountainside 37.5 s after firing. What are the x and y components of the shell where it explodes, relative to its firing point?
x = __________m
y = __________m
To solve the problem, we need to find the horizontal (x) and vertical (y) components of the shell's position at the time it explodes.
First, let's break down the given information:
- Initial velocity of the shell: 300 m/s
- Launch angle above the horizontal: 65.0°
- Time elapsed before the shell explodes: 37.5 s
We'll use the following kinematic equations to find the x and y components of the position:
1. Vertical position equation:
y = y0 + v0y * t - (1/2) * g * t^2
2. Horizontal position equation:
x = x0 + v0x * t
Now, let's calculate the components:
Step 1: Find the initial x and y components of the velocity (v0x and v0y):
v0x = v0 * cos(θ)
v0y = v0 * sin(θ)
where θ is the launch angle in radians (θ = 65.0° * π/180).
v0x = 300 m/s * cos(65.0° * π/180)
v0y = 300 m/s * sin(65.0° * π/180)
Step 2: Calculate the initial position (x0 and y0):
Assuming the shell is fired from the origin (0, 0), so x0 = 0 and y0 = 0.
Step 3: Calculate the final horizontal and vertical positions (x and y):
Substitute the values into the position equations mentioned earlier.
x = x0 + v0x * t
y = y0 + v0y * t - (1/2) * g * t^2
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Step 4: Plug in the values to find the x and y coordinates where the shell explodes:
x = v0x * t
y = v0y * t - (1/2) * g * t^2
x = 300 m/s * cos(65.0° * π/180) * 37.5 s
y = 300 m/s * sin(65.0° * π/180) * 37.5 s - (1/2) * 9.8 m/s^2 * (37.5 s)^2
Now, calculate the values to find x and y.
a = V / t,
a = 300 / 37.5 = 8 m/s^2.
d = Vo*t + 0.5at^2,
d = 300 * 37.5 + 0.5 * 8 * (37.5)^2,
= 11250 + 5625 = 16875 m @ 65 deg.
X = hor = 16875 * cos65 = 7132 m.
Y = ver = 16875 * sin65 = 15294 m.