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maks197457 [2]
4 years ago
13

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.
Physics
1 answer:
Basile [38]4 years ago
7 0

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).

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