Can Quentin Hit the Bullseye? Calculate Here!

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically analyzing whether a dart thrown horizontally can hit a bullseye on a dartboard from a distance of 2.0 meters. The original poster presents the scenario with specific parameters, including the speed of the dart and the dimensions of the bullseye.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss using basic kinematic equations to determine the time of flight and the vertical distance fallen by the dart. There are questions about the validity of the calculations and the assumptions made regarding the time of flight.

Discussion Status

Some participants have provided guidance on the calculations, while others have pointed out errors in the original calculations. There is an ongoing exploration of the implications of these calculations on whether Quentin can hit the bullseye.

Contextual Notes

Participants note that the original calculations led to an implausible result, prompting a reevaluation of the time of flight and the method used. There is an acknowledgment of the need to clarify the assumptions regarding the dart's trajectory and the initial conditions of the throw.

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Quentin claims that he can throw a dart at dartboard from distance of 2.0 m & hit 5 cm wide bullseye if he throws dart horizantally with speed of 15 m/s. He starts the throw at the same height as top of bullseye. See if Quentin is able to hit bullseye by calculating how far his shot falls from the bullseye's lower edge. THANKS IN ADVANCE!
 
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Here's a way:
Use [tex]d=vt[/tex] to find the amount of time the dart flies.
Then use [tex]h=1/2gt^2[/tex] to find what distance it has fallen.
 
Dart problem

Thanks Galileo for your suggestion. I should have posted that I tried that and what I get is:

using x=vt, 2=15t, t=7.5 secs

y at 7.5 secs would be (using 1/2gt^2): -4.9 X 56.25 = -275.625

275.625 m below the throwing height doesn't make sense to me or does it mean that the answer is: Quentin will not be able to hit the bullseye. Thanks
 
The time of flight you calculated is wrong.
It is actually much lower.
If 7.5 sec was correct, Quentin could have run faster than the dart.
 
Dart problem

Thank you alalbatros! I was dividing 15 by 2 instead of the other way around. It makes sense now. Thanks again.
 

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