An archer shoots an arrow horizontally at a target 12 m away. The arrow is aimed directly at the center of the target, but it hits 59 cm lower. What was the initial speed of the arrow?
Added by Brenna S.
Step 1
59 m), \(h\) is the initial height of the arrow, \(g\) is the acceleration due to gravity (9.8 m/s\(^2\)), and \(t\) is the time of flight. Show more…
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An archer shoots an arrow horizontally at a target 12 m away. The arrow is aimed directly at the center of the target, but it hits 20 cm lower. What is the initial speed of the arrow? Given: Distance to the target (d) = 12 m Vertical displacement (h) = 20 cm = 0.2 m Required: Initial speed of the arrow (v) Solution: We can use the kinematic equation for vertical motion to solve for the initial speed of the arrow. The equation is: h = (1/2) * g * t^2 Where: h = vertical displacement g = acceleration due to gravity (approximately 9.8 m/s^2) t = time of flight Since the arrow is shot horizontally, the time of flight (t) can be calculated using the horizontal distance (d) and the initial speed (v): t = d / v Substituting this into the kinematic equation, we have: 0.2 = (1/2) * 9.8 * (d / v)^2 Simplifying, we get: 0.2 = 4.9 * (d^2 / v^2) Rearranging the equation, we can solve for v: v^2 = (d^2 * 4.9) / 0.2 Taking the square root of both sides, we find: v = sqrt((d^2 * 4.9) / 0.2) Now we can substitute the given values and calculate the initial speed of the arrow: v = sqrt((12^2 * 4.9) / 0.2)
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