Question

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

Answer 1

(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;

$t = \frac{X}{v} \\\\t = \frac{20 }{80} \\\\t = 0.25 \ s$

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

$h = v_0_y t + \frac{1}{2} gt^2\\\\h = 0 \ + \ \frac{1}{2} \times 9.8 \times 0.25^2\\\\h =0.31 \ m$

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