An archer shoots an arrow at 63.2m/s at a 42.0 angle. if the ground is flat, how far away does it land(unit m)

  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.

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.

the force represented by vector a is normal force. if the hockey puck is moving to the right, when it is hit with a significant force by the hockey stick the puck will move off in a different direction. correct on edge.