Projectile Motion velocity and direction

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Homework Help Overview

The problem involves projectile motion, specifically analyzing the trajectory of a projectile launched from the origin with an initial velocity and angle, and determining its intersection with a non-linear ground surface described by a quadratic equation.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the challenge of dealing with a non-constant ground function compared to simpler cases with flat ground. There is confusion regarding the correct application of motion equations and the placement of powers in the equations.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the ground function on the projectile's trajectory. Some guidance has been provided regarding the equations of motion, but there is no clear consensus on the next steps or resolution of the confusion.

Contextual Notes

Participants are grappling with the complexity introduced by the quadratic ground profile and its effect on determining the projectile's impact point and motion characteristics.

aaronfue
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Homework Statement



At t = 0, a projectile is located at the origin and has a velocity of 20 m/s at 40° above the horizontal. The profile of the ground surface it strikes can be approximated by the equation y = 0.4x – 0.006x2, where x and y are in meters. Determine (a) the approximate coordinates of the point where it hits the ground, (b) its velocity and direction when it hits the ground, (c) its highest point in flight and (d) the greatest distance above the ground.

Homework Equations



vo= 20 \frac{m}{s}
θ=40°
Horizontal Motion: x=xo2 + (vo)x2t + \frac{1}{2}at
Vertical Motion: y=yo2 + (vo)y2t + \frac{1}{2}at

The Attempt at a Solution



This function of the ground is confusing me. I'm not sure where to start with this.
 

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When you do one of these problems with a flat ground you also have do deal with a function that describes the height of the ground at a given distance, but it is easy since the function is constant. Here you do the same thing, but you have a non-constant function for the ground. In both cases you find the place where the trajectory intersects the ground.
 
aaronfue said:
Horizontal Motion: x=xo2 + (vo)x2t + \frac{1}{2}at
Vertical Motion: y=yo2 + (vo)y2t + \frac{1}{2}at
The powers of 2 are in the wrong places.
 
haruspex said:
The powers of 2 are in the wrong places.

Got it. Thanks!
 

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