At takeoff, the horizontal and vertical velocity of long jumper are 57.8 m/s and 5.6 m/s respectively. what is the resultant velocity

  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.

easy math just plug in a and b

step-by-step explanation:

(23*2)-12=x     x=34

step-by-step explanation: