Angular Velocity: Find Launch Angles for Projectile

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SUMMARY

The discussion focuses on calculating the launch angles for a projectile that travels 100 meters horizontally while just clearing a 10-meter wall, with an initial velocity of 36 m/s. Participants suggest using the equations of motion, specifically the constant acceleration equations for both the horizontal and vertical components. By eliminating time from these equations, a quadratic equation in terms of the launch angle θ can be derived, yielding two distinct solutions for the angle of launch.

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  • Understanding of projectile motion and its equations
  • Familiarity with constant acceleration equations
  • Knowledge of trigonometric functions and their applications in physics
  • Ability to solve quadratic equations
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  • Learn how to apply trigonometric identities in physics problems
  • Explore the concept of maximum height in projectile motion
  • Practice solving quadratic equations in the context of physics applications
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trigunesq
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Homework Statement


A projectile is launched at 36 m/s and travels 100 m horizontally. If it is to be able to just barely clear a 10 m wall, between what two angles should the projectile be launched with respect to the horizontal?




Homework Equations


vf=vo+at
x=volt+1/2at^2
vf^2=Vo^2+2ax
 
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Welcome to PF!

Hi trigunesq! Welcome to PF! :smile:

(have a theta: θ and try using the X2 and X2 tags just above the Reply box :wink:)

They're asking you for the two angles from which it will be 10m up when it's 100m along.

Call the launch angle θ, and the time taken t, then use your constant acceleration equations for the x and y directions separately (a = 0 for the direction, of course), to get a pair of equations from which you can eliminate t to get a quadratic equation for θ (with two solutions! :wink:).

What do you get? :smile:
 

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