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Carnegie Mellon University
Problem 16 Easy Difficulty
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?
Answer
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.
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Top Physics 101 Mechanics Educators
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Video Transcript
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.
Carnegie Mellon University
Topics
Motion Along a Straight Line
Motion in 2d or 3d
Newton's Laws of Motion
Rotation of Rigid Bodies
Dynamics of Rotational Motion
Equilibrium and Elasticity
Top Physics 101 Mechanics Educators
Andy C.
University of Michigan - Ann Arbor
Marshall S.
University of Washington
Farnaz M.
Other Schools
Jared E.
University of Winnipeg
