Like

Report

An artillery shell is fired with an initial velocity of 300 $\mathrm{m} / \mathrm{s}$ at $55.0^{\circ}$ above the horizontal. To clear an avalanche, it explodes on a mountainside 42.0 $\mathrm{s}$ after firing. What are the $x-$ and $y$ -coordinates of the shell where it explodes, relative to its firing point?

The shell is in $x = 7227.06 \mathrm { m }$ and $\mathrm { y } = 1677.72 \mathrm { m }$

That's at a horizontal distance of 7227.06$\mathrm { m }$ from firing point and at a height of

1677.72$\mathrm { m }$ from firing point.

You must be signed in to discuss.

University of Michigan - Ann Arbor

University of Washington

Other Schools

University of Winnipeg

for the exposition, we could say that displacement in the X on the extraction would be equaling velocity ex initial times t Because there isn't any. There isn't any acceleration in the ex direction. So this would be 300 meters per second multiplied by co signed 55 degrees multiplied by 40 2.0 seconds. This is equaling essentially 700 7000. Rather 200 7000 220 seven meters for Delta. Why, this would be equally the y initial T plus 1/2 g t squared. We're gonna choose positive. We're gonna choose upwards to be positive. And so delta Y is equaling 300 meters per second sign of 55 degrees times 42.0 seconds plus 1/2 times negative, 9.80 meters per second squared times 42.0 seconds. Quantity squared and we find that delta Y is equaling here. It would be 1678 meters, so this would be your position. Your final positions for your final coordinates would be 7227 meters comma 1678 meters. This would be your final position in your final answer. That is the end of the solution. Thank you for watching.