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

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SUMMARY

The discussion focuses on determining the launch angle (theta) for a projectile to achieve specific horizontal (2000m) and vertical (800m) distances. The initial velocity is set at 1000m/s. The user calculated one solution at approximately 22.4 degrees but is confused about the second solution of 89.4 degrees provided in the answer key. The correct approach involves using the projectile motion equations, specifically y = x tan(theta) - (gx^2)/(2u^2 cos^2(theta)), to derive both angles accurately.

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  • Knowledge of kinematic equations, particularly for vertical and horizontal motion
  • Ability to manipulate algebraic equations to solve for unknowns
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  • Study the derivation of the projectile motion equations, focusing on the formula y = x tan(theta) - (gx^2)/(2u^2 cos^2(theta)
  • Learn how to apply trigonometric identities to solve for multiple angles in projectile motion
  • Explore numerical methods for solving nonlinear equations to find additional solutions
  • Review examples of projectile motion problems with varying launch angles and initial velocities
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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 - \stackrel{gx^2}{2u^2cos^2 theta}

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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