Finding the initial velocity using quadratic

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SUMMARY

The discussion focuses on calculating the initial velocity (Vi) of a projectile using kinematic equations and quadratic formulas. The user initially derived an equation involving Vi, time (Δt), and vertical displacement (Δy) but encountered difficulties in solving for Vi. After an attempt, the user calculated Vi as 9.9 m/s, while the correct answer, as indicated by the teacher, is 17.05 m/s. The discrepancy suggests an algebraic error in the user's calculations.

PREREQUISITES
  • Understanding of kinematic equations, specifically Δy=ViyΔt + 1/2ayΔt^2
  • Familiarity with projectile motion concepts, including initial velocity and angle of projection
  • Knowledge of quadratic equations, particularly the formula x= [-b +/- √(b^2-4ac)]/2a
  • Basic trigonometry, including sine and cosine functions for angle calculations
NEXT STEPS
  • Review the derivation of kinematic equations in projectile motion
  • Practice solving quadratic equations with real-world physics problems
  • Explore the impact of angle on projectile motion using simulation tools
  • Learn how to correctly apply trigonometric functions in physics problems
USEFUL FOR

Students studying physics, particularly those focusing on kinematics and projectile motion, as well as educators looking for examples of common algebraic errors in problem-solving.

Rammer24
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Homework Statement



I need to find Vi, knowing the following:

Δt=6/Vi cos40

Viy=Vi sin40

Δy=4m

ay=-g = -9.8m/s^2

Homework Equations



Quadratic: x= [-b +/- √(b^2-4ac)]/2a

Kinematic equation to be used: Δy=ViyΔt + 1/2ayΔt^2

The Attempt at a Solution



Δy=ViyΔt + 1/2ayΔt^2

4=(Vi sin40)(6/Vi cos40) -4.9(6/Vi cos40)^2

This is where I'm stuck, I don't know how to continue the problem and solve for Vi.

EDIT: I solved for Vi and got and answer of 9.9m/s, but according to my teacher, the correct answer should be 17.05m/s, so I'm definitely making a mistake somewhere in the algebra.

Thanks
 
Last edited:
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