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An arrow is shot at a target 20 m away. The arrow is shot with a horizontal velocity of 80 m/s.

How long is the arrow in the air for?____s

How far does the arrow drop as it approaches the target?____m

Sagot :

onyebuchinnaji

(1) The time of motion of the arrow is 0.25 s.

(2) The vertical height dropped by the arrow as it approaches the target is 0.31 m.

The given parameters:

  • Horizontal distance of the arrow, X = 20 m
  • Horizontal speed of the arrow, v = 80 m/s

The time of motion of the arrow is calculated as follows;

[tex]t = \frac{X}{v} \\\\t = \frac{20 }{80} \\\\t = 0.25 \ s[/tex]

The vertical height dropped by the arrow as it approaches the target is calculated as follows;

[tex]h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0 \ + \ \frac{1}{2} \times 9.8 \times 0.25^2\\\\h =0.31 \ m[/tex]

Learn more about time of motion of projectile here:  https://brainly.com/question/1912408

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