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In a live fire exercise, an Army artillery team fires an artillery shell from a howitzer. The barrel of the howitzer makes a 60.0° angle above horizontal, and the speed of the shell upon exiting the barrel is 380 m/s. The shell hits a target on the side of a mountain 32.0 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.

Respuesta :

Answer:

Explanation:

speed of shell =380 m /s

x - component = 380 cos 60

= 190 m /s

y- component = 380 sin 60

= 329 .1 m /s

time taken to hit target = 32 s

horizontal distance covered = horizontal velocity x time

= 190 x 32 = 6080 m

vertical distance covered

h = ut - 1/2 gt²

=  329.1 x 32 - .5 x 9.8 x 32²

=  10531.2 - 5017.6

= 5513.6 m .

x coordinate = 6080 m

y-coordinate = 5513.6 m

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