Projectile Problem: Distance and Max Height

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SUMMARY

The projectile problem involves a projectile launched at a 45-degree angle with an initial velocity of 100√2 ft/sec, following the trajectory described by the equation y = x - (x/25)². The horizontal distance traveled before hitting the ground is determined to be 50 ft, while the maximum height attained by the projectile is 25 ft. The vertex of the parabola, representing the highest point, can be found using the formula for the vertex of a quadratic equation.

PREREQUISITES
  • Understanding of projectile motion principles
  • Familiarity with quadratic equations and parabolas
  • Knowledge of vertex calculation for parabolas
  • Basic trigonometry, specifically angle measurements
NEXT STEPS
  • Study the derivation of projectile motion equations
  • Learn how to calculate the vertex of a parabola in detail
  • Explore the effects of different launch angles on projectile distance and height
  • Investigate real-world applications of projectile motion in physics
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Students studying physics, educators teaching kinematics, and anyone interested in understanding the mathematics of projectile motion.

GusTu2007
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Suppose that a projectile is fired at an angle of 45[tex]\circ[/tex] from the
horizontal. Its initial position is the origin in the xy-plane, and its
initial velocity is 100[tex]\sqrt{}2[/tex] ft / sec. Then its trajectory will be the part of
the parabola y = x -( [tex]\frac{}{}x/[/tex]25)^2 for which y ≥ 0.
a) How far does the projectile travel (horizontally) before it hits the
ground?
b) What is the maximum height above the ground that the
projectile attains?
 
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a) When it hits the ground again, what's the value of y?
b) The highest point is at the vertex of the parabola. Do you know how to find the vertex from the given eqn?
 

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