Based on the height of the marble launcher above the floor and the initial vertical velocity component, calculate the amount of time needed for the projectile to reach the ground. (Note that this calculation will require you to solve a quadratic equation. Ask your teacher if you need help with this calculation.) angle 1: θ = 15° -->initial height 0.027m, horizontal distance 0.145m, vertical velocity component 1.12m/s angle 2: θ = 45° -->initial height 0.027m, horizontal distance 0.209m, vertical velocity component 3.05m/s angle 3: θ = 70° --> initial height 0.027m, horizontal distance 0.118m, vertical velocity component 4.05m/s

Based on the height of the marble launcher above the floor and the initial vertical velocity component, calculate the amount of time needed for the projectile to reach the ground. (Note that this calculation will require you to solve a quadratic equation. Ask your teacher if you need help with this calculation.) angle 1: θ = 15° -->initial height 0.027m, horizontal distance 0.145m, vertical velocity component 1.12m/s angle 2: θ = 45° -->initial height 0.027m, horizontal distance 0.209m, vertical velocity component 3.05m/s angle 3: θ = 70° --> initial height 0.027m, horizontal distance 0.118m, vertical velocity component 4.05m/s

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Description: Based on the height of the marble launcher above the floor and the initial vertical velocity component, calculate the amount of time needed for the projectile to reach the ground. (Note that this calculation will require you to solve a quadratic equ

Physics for Scientists and Engineers: Foundations and Connections

1st Edition

ISBN:9781133939146

Author:Katz, Debora M.

Publisher:Katz, Debora M.

Chapter3: Vectors

Section: Chapter Questions

Problem 42PQ: The same vectors that are shown in Figure P3.6 are shown in Figure P3.42. The magnitudes are F1 =...

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Based on the height of the marble launcher above the floor and the initial vertical velocity component, calculate the amount of time needed for the projectile to reach the ground. (Note that this calculation will require you to solve a quadratic equation. Ask your teacher if you need help with this calculation.)

angle 1: θ = 15° -->initial height 0.027m, horizontal distance 0.145m, vertical velocity component 1.12m/s

angle 2: θ = 45° -->initial height 0.027m, horizontal distance 0.209m, vertical velocity component 3.05m/s

angle 3: θ = 70° --> initial height 0.027m, horizontal distance 0.118m, vertical velocity component 4.05m/s

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