Answer:
58.071 m/s @ 5.53°
Step-by-step explanation:
The magnitude of the velocity vector can be found using the Pythagorean theorem:
|v| = √(57.8² +5.6²) m/s ≈ 58.071 m/s
The angle of the velocity vector with respect to the horizontal can be found using the arctangent function:
∠v = arctan(5.6/57.8) ≈ 5.53°
So, the velocity vector is ...
58.071∠5.53° m/s
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In the attached graph, the vertical scale is exaggerated, so the take-off angle looks to be greater than it actually is. "Hang time" is about 1143 milliseconds.
