An artillery shell is fired with an initial velocity of 300m/s at 55 degrees above horizontal. To clear an avalanche, it explodes on a mountainside 42.0s after firing. What are the x and y coordinates of the shell where it explodes, relative to its firing point?
1.An artillery shell is fired with an initial velocity of 300 m/s at 55° above the
a)Determine its maximum height
b)How long does it take to reach the maximum height
horizontal. It explodes on the same ground level.
c)Determine the range of its flight
7227.1
To find the x and y coordinates of the shell where it explodes, relative to its firing point, we can break down the problem into horizontal and vertical components.
First, let's find the horizontal displacement (x-coordinate).
The initial velocity of the shell can be divided into horizontal (Vx) and vertical (Vy) components using trigonometry:
Vx = V * cos(θ)
Vx = 300 m/s * cos(55°)
Now, we can calculate the horizontal displacement using the equation:
x = Vx * t
x = (300 m/s * cos(55°)) * 42.0 s
Next, let's find the vertical displacement (y-coordinate).
The vertical component of the initial velocity can be calculated as:
Vy = V * sin(θ)
Vy = 300 m/s * sin(55°)
To find the vertical displacement, we'll use the equation:
y = (Vy * t) + (0.5 * g * t^2)
Since we need the displacement at the time of explosion, we can substitute the given time (42.0 s) into the equation.
y = (300 m/s * sin(55°) * 42.0 s) + (0.5 * 9.8 m/s^2 * (42.0 s)^2)
Now, we can calculate the values of x and y to find the coordinates where the shell explodes.
Hor vel= 300cos55
Mult this by 42s to get the horizontal (x) distance. (42s btw is a bit ridiculous, but hey ho)