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= ?
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
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
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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