Answer:
resolving the velocity into its components (the vertical and horizontal direction)
Uy = 63.2sin42
Ux = 63.2cos42
the acceleration in the X direction is 0 while in the Y direction is 9.82 m/s²
apply the 3 equations of motion with constant acceleration in both directions
to find how far the arrow goes horizontally, first calculate the time taken by it to reach its highest point then hit the ground
Y = Uy + 0.5gt² (put y = 0 because the height of the archer wasn't given)
0 = 63.2sin42t - 0.5*9.8*t²
t = 8.6
then apply it in X direction
X = 63.2cos42 * 8.6 - 0 = 403 meters
Answer:
405.3 m
Explanation:
The initial vertical speed is ...
Vv = (63.2 m/s)sin(42°) ≈ 42.2891 m/s
The acceleration due to gravity reduces this to zero in ...
t = Vv/(-g) = (42.2891 m/s)/(9.8 m/s²) ≈ 4.31521 s
This is the time it takes for the arrow to reach its highest point. The time it takes to fall back to earth will be the same, so the total flight time will be ...
2×4.31521 s ≈ 8.63042 s
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The horizontal velocity is ...
Vh = (63.2 m/s)cos(42°) ≈ 46.9668 m/s
In the time of flight, the horizontal distance covered is ...
(8.63042 s)(46.9668 m/s) ≈ 405.343 m
The arrow lands approximately 405.3 meters away.
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Comment on the solution
This solution assumes the arrow was launched from ground level. Its flight distance will be farther if it was launched from a height above ground level.
If you work the problem symbolically, you find the solution to be ...
d = v²·sin(2θ)/g = (63.2 m/s)²/(9.8 m/s²)·sin(84°) = 405.343 m
In the above, we have worked the problem using the gravitational acceleration usually used in math problems. The standard value for g is 9.80665. Using that gives a distance of 405.068 m. Either answer rounds to 405 m.
