What Angle Should a Projectile Be Launched to Reach Specific Distances?

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Homework Help Overview

The discussion revolves around a projectile motion problem where a ball is launched at a specific velocity and must reach designated horizontal and vertical distances. Participants are exploring the angle of launch required to achieve these distances.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to find the launch angle using kinematic equations and expresses confusion over the existence of two solutions for the angle. Other participants discuss the use of a specific formula and share their own results, questioning the validity of the answer key.

Discussion Status

Participants are actively engaging with the problem, sharing their calculations and questioning the provided solutions. There is recognition of the possibility of multiple angles, but no consensus on how to derive the second angle has been reached.

Contextual Notes

Some participants note that they are unsure how to derive the second angle and express uncertainty about the accuracy of the answer key, indicating potential gaps in information or understanding.

clipperdude21
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Projectile Motion PLEASE HELP MIDTERM TOMMOROW!

1. A ball is launched with a velocity of 1000m/s at some angle theta. The ball must travel 2000m in the x dir and land 800m in the y dir on a plateau. What theta should the ball be shot at to make this happen.



2. y=y0 +v0t -.5gt^2



3. I solved for t using x=v0cos(theta)t and plugged that into the y equation and got one solution. The answer key says there's two solutions. One at 20 something, which is the one i got and also one at 89.4 degrees. Can someone show me how to get the other/both solutions


THANKS! I appreciate it
 
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You should use the formula
y=x tan theta - [tex]\stackrel{gx^2}{2u^2cos^2 theta}[/tex]

THis formula is combined by
verticial component using s=ut+1/2 at^2
horizontal component v=s/t
 
i used that equation and still only got 22.4 degrees at the sole solution... 89.4 doesn't seem to fit. Is there an error in the answer key?
 
i also get the 22.4 degree
but it is acceptable that there is another angle is the answer...
however I forget how to solve it ~sorry
 
ok thanks anyway!
 

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