Projectile Motion of Dart: Calculating Distance and Release Point

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The discussion focuses on calculating the distance PQ and the release point of a dart thrown horizontally at 11 m/s, which hits the board 0.19 seconds later. Part (a) involves determining the vertical distance the dart falls during its flight, confirming that it relates to vertical motion. The formula discussed for calculating the y-coordinate at impact is y = 11sin(0) * 0.19 - 4.9 * (0.19)^2, which is correct for this scenario. The participants clarify the relationship between horizontal speed and vertical displacement in projectile motion. Understanding these calculations is essential for accurately determining the dart's trajectory and impact point.
frankfjf
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A dart is thrown horizontally with an initial speed of 11 m/s toward point P, the bull's-eye on a dart board. It hits at point Q on the rim, vertically below P, 0.19 s later. (a) What is the distance PQ? (b) How far away from the dart board is the dart released?

Is part a basically asking for the vertical motion?

Thanks.
 
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frankfjf said:
Is part a basically asking for the vertical motion?
Right. In other words: How far does the dart fall during its flight?
 
Just confirming that my formula is right here, wouldn't that be obtained by:

y = 11sin(0).19 - 4.9(.19)^2?
 
Right. That gives the y-coordinate of the dart when it hits the board.
 

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