An artillery shell is fired with an initial velocity of 300 m/s at 64.0° above the horizontal. To clear an avalanche, it explodes on a mountainside 48.0 s after firing. What are the x- and y-coordinates of the shell where it explodes, relative to its firing point?x= ?y= ?

Question
Asked Nov 21, 2019
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An artillery shell is fired with an initial velocity of 300 m/s at 64.0° above the horizontal. To clear an avalanche, it explodes on a mountainside 48.0 s after firing. What are the x- and y-coordinates of the shell where it explodes, relative to its firing point?

x= ?

y= ?

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Expert Answer

Step 1

Given:

The initial velocity, u = 300 m/s
Angle, 0 640
Velocity along x-axis, u. = ucos0 300 cos640 = 131.5 1m/ s
Velocity along y-axis, u, = usin 64° = 300 sin 64° = 269.64m /s
Time of travel, t = 48s
Along the X-coordinate, the velocity will be constant and along the Y-coordinate
the velocity (y ) will have deacceleration due to gravity
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The initial velocity, u = 300 m/s Angle, 0 640 Velocity along x-axis, u. = ucos0 300 cos640 = 131.5 1m/ s Velocity along y-axis, u, = usin 64° = 300 sin 64° = 269.64m /s Time of travel, t = 48s Along the X-coordinate, the velocity will be constant and along the Y-coordinate the velocity (y ) will have deacceleration due to gravity

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Step 2

X-coordinate

After 48s, the x-coor dinate (x), where it explodes will be:
Distance Velocity x Time [as there is no acceleration]
(131.5m/ s)(48s)
= 6312.48m
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After 48s, the x-coor dinate (x), where it explodes will be: Distance Velocity x Time [as there is no acceleration] (131.5m/ s)(48s) = 6312.48m

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Step 3

Y-coordinate has a negative acce...

The y-coordinate (y), according to second equati on of motion
2
(9.81)(48)
- (269.64m / s)(48s)-
2
- (12942.72)-11301.12)
=1641.6m
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The y-coordinate (y), according to second equati on of motion 2 (9.81)(48) - (269.64m / s)(48s)- 2 - (12942.72)-11301.12) =1641.6m

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