Projectile Motion, Vector Component Confusion

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SUMMARY

The discussion centers on calculating the time a projectile, fired at 300 m/s at a 45-degree angle, remains in the air under the influence of gravity (-9.8 m/s²) without air resistance. The initial horizontal and vertical velocity components are determined as ViX = 212 m/s and ViY = 212 m/s, respectively. The maximum displacement reached is 2293 m, calculated using the formula vf² = vi² + 2a(d). The time of flight is derived from the vertical motion equations, specifically using the condition that vertical velocity (Vy) equals zero at the peak of the trajectory.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with trigonometric functions (sine and cosine)
  • Knowledge of kinematic equations for motion
  • Basic algebra skills for rearranging equations
NEXT STEPS
  • Study the derivation of the time of flight formula for projectile motion
  • Learn how to apply the kinematic equation dy = vi(t) + 1/2 a(t)² effectively
  • Explore the impact of air resistance on projectile motion
  • Investigate the use of vector components in two-dimensional motion analysis
USEFUL FOR

Students in physics, educators teaching projectile motion concepts, and anyone interested in mastering kinematic equations and vector analysis in motion problems.

cire792
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Homework Statement



Canon fired at 45Degrees, at 300m/s
Assuming no air resistance,
a= -9.8m/s^2
ViX = 300cos45 = 212m/s
ViY = 300sin45 - 212m/s
Displacement = 2293m [used vf^2=vi^2+2a(d)] rearranged for d, and vf=0, at highest peak.

Time the ball is in the air= ?

Homework Equations



Thinking I have to use this, but this is where my math skills are rusty.

dy = vi (t) + 1/2 a(t)^2

The Attempt at a Solution



time = -1038
 
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You noted that Vy = 0 at half time.
So you can use Vy = Viy + at to find the half time. Viy = 212
 

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