Solving 2D Kinematics Problem: Distance OP in m

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The problem involves calculating the vertical distance OP that an arrow falls in 0.10 seconds while shot horizontally towards point O, 20.0 meters away. The arrow experiences free fall with an acceleration of 9.8 m/s². Using the kinematic equation x = 0.5*a*t² + v*t, where the initial vertical velocity (v) is zero, the distance OP can be calculated as 0.5 * 9.8 * (0.10)², resulting in a distance of 0.049 meters.

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I have a problem with this question.
An arrow is shot horizontally towards point O, which is at a distance of 20.0 m. It hits point P (right below point O) 0.10 s later. What is the distance OP (in metres)?
What do i need to find? How would i know the time taken for the arrow to reach o?
 
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The distance to the target is a red herring; this is a question of kinematics in 1D: the vertical dimension.

You know the arrow has been in free fall for 0.10s, and when something is in free-fall it accelerates downwards at 9.8 m/s^2. Since the arrow was shot horizontally, it's initial downward velocity is zero, and you can use

x = 0.5*a*t^2 + v*t

to find your distance.
 

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