For safety reasons, park rangers decide to start an avalanche on a mountain slope. They fire an artillery shell at an angle of 430 above the horizontal with an initial speed of 284 m/s. Thirty seconds later they see the explosion. What is the x coordinates of the shell where it explodes, relative to the firing point?
/i havent the slightest on how to start
the horizontal (x) component of the launch speed is ... 284 m/s * cos(43º)
the horizontal component is not affected by gravity
... so just multiply by the flight time
That did the trick!
I see now, we're rolling in trig functions!
math is not my strong point, thank you again Scott.
To find the x-coordinate where the shell explodes, we need to consider the horizontal motion of the shell.
First, let's break down the given information:
Angle of projection (θ) = 43° above the horizontal
Initial speed (v0) = 284 m/s
Time of flight (t) = 30 seconds
We know that the horizontal and vertical motions are independent of each other. The horizontal motion can be described using the equation:
x = v0 * cos(θ) * t
In this equation, x represents the horizontal distance traveled, v0 is the initial speed, θ is the launch angle, and t is the time of flight.
To calculate the x-coordinate where the shell explodes, we substitute the given values into the equation:
x = 284 m/s * cos(43°) * 30 s
Now, let's calculate it step by step:
1. Convert the angle from degrees to radians:
θ_radians = 43° * (π/180°)
2. Calculate the cosine of the angle:
cos(θ_radians) = cos(43°)
3. Substitute the values into the equation:
x = 284 m/s * cos(43°) * 30 s
Now, let's calculate it using a calculator:
θ_radians ≈ 0.749 radians
cos(θ_radians) ≈ 0.7314
x = 284 m/s * 0.7314 * 30 s
x ≈ 6267 meters
Therefore, the x-coordinate of the shell where it explodes, relative to the firing point, is approximately 6267 meters.