Stone projectile motion problem

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SUMMARY

The stone projectile motion problem involves a stone projected from ground level at an initial speed of 80 m/s, with a peak speed of 60 m/s. The angle of projection can be calculated using the formula v0x = v0 * cos(theta), where v0x is the horizontal component of velocity. The correct angle of projection is determined to be approximately 38.9 degrees. The discussion highlights the importance of understanding the relationship between horizontal and vertical components of projectile motion, particularly in the absence of wind resistance.

PREREQUISITES
  • Understanding of basic projectile motion concepts
  • Familiarity with trigonometric functions, specifically cosine
  • Experience with using scientific calculators, such as the TI-89
  • Knowledge of the equations of motion in physics
NEXT STEPS
  • Learn how to derive projectile motion equations in physics
  • Study the effects of air resistance on projectile motion
  • Explore advanced trigonometric identities and their applications
  • Practice solving various projectile motion problems with different initial conditions
USEFUL FOR

This discussion is beneficial for physics students, educators, and anyone interested in mastering projectile motion calculations and understanding the underlying principles of motion in a vacuum.

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A stone is projected from ground level at 80 m/s. Its speed when it reaches its highest point is 60 m/s.

Find the angle of its projection. Answer in units of degrees.

Lost count how many times i tried to get this one right.
 
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Post your working, so we can show you where you've gone wrong.
 
OMG! i did it...it was my calculators fault! can't believe this.

here try this if you have a ti-89 (set mode to degrees)
(note the real values of the problem were 56m/s and 72m/s)

solve(56 = 72 * cos(x), x)

the answer it gives is weird

x = 360. * ((weird symbole)n8 + .108173447969) or x = 360. * ((weird symbole)n8 - .108173447969)

everything in my work made sense so i tried to do that manually:

56/72 = 72/72 * cos(x)
.77778 = cos(x)
cos-1(.77778) = x
x = 38.9


Now the weird thing...putting this in calculator backwards with the value i determiend for theta:

72 * cos(38.9) gets me 56! Freekin modern technology lol

o yea if anyone else wants to know how to do this problem, remember that in problems they neglect wind resistance the horizontal velocity is constant...so at the projectiles heighest point its velocity in the y direction is zero, so the velocity in the x direction is its total velocity...so now you have v0x...and they gave v0 so the formula v0x = v0 * cos(theta) will get you theta...i just figured this out now because i wanted to organize my messed up attempts for posting :P
 
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