The time taken for the arrow to hit the ground is 0.53 s.
The horizontal distance traveled by the arrow is 25.97 m.
The given parameters:
- Initial horizontal velocity, [tex]v_x[/tex] = 49 m/s
- Initial vertical velocity, [tex]v_y[/tex] = 0
What is time of motion?
- This is the time taken for a projectile to land on the plane of projection.
The time of motion is calculated as follows;
[tex]h = v_yt + \frac{1}{2} gt^2\\\\h = 0 + \frac{1}{2} gt^2\\\\h = \frac{1}{2} gt^2\\\\t = \sqrt{\frac{2h}{g} } \\\\t = \sqrt{\frac{2\times 1.4}{9.8} }\\\\t = 0.53 \ s[/tex]
The horizontal distance traveled by the arrow is calculated as follows;
[tex]X = v_x t\\\\X = (49\ m/s) \times (0.53 \ s)\\\\X = 25.97 \ m[/tex]
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