Question

In a live fire exercise, an Army artillery team fires an artillery shell from a howitzer. The barrel of the howitzer makes a 50.0° angle above horizontal, and the speed of the shell upon exiting the barrel is 360 m/s. The shell hits a target on the side of a mountain 38.0 s after firing. Assuming the point where the shell exits the barrel to be the origin, and assuming as usual that the x-axis is horizontal and the y-axis is vertical, find the x and y coordinates, in meters, of the target.

Answer 1

Answer:

The coordinates of the target are (8790m, 3400m).

Explanation:

First of all, we have to find the components of the initial velocity $v_0_x$ and $v_0_y$, using trigonometry:

$v_0_x=v_0\cos\theta=(360m/s)\cos50\°=231.4m/s\\\\v_0_y=v_0\sin\theta=(360m/s)\sin50\°=275.7m/s$

Now, we find the x-coordinate using the equation of motion with constant speed (since there is no external force in x-axis that causes an horizontal acceleration):

$x=v_0_xt\\\\x=(231m/s)(38.0s)=8790m$

Then, we find the y-coordinate using the equation of position of an object with constant acceleration (since there is the gravitational force causing a vertical acceleration on the shell):

$y=v_0_yt-\frac{1}{2}gt^{2}\\\\y=(276m/s)(38.0s)-\frac{1}{2}(9.81m/s^{2})(38.0s)^{2}=3400m$

Finally, the coordinates of the target are (8790m, 3400m).