Projectile Motion Homework: Find y(x) & Graph -45, 45 deg

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Homework Help Overview

The problem involves analyzing the trajectory of a rocket-powered hockey puck moving on frictionless ice, specifically focusing on its motion along the y-axis while thrust is applied at an angle. The task includes deriving an algebraic equation for the puck's trajectory and graphing its motion at angles of 45 and -45 degrees.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the equations of motion and attempt to derive the trajectory equation y(x). There are questions about the effects of thrust and gravity on the puck's motion at different angles. Some participants explore the relationship between x and y accelerations and their dependence on the angle of thrust.

Discussion Status

The discussion includes various attempts to derive the correct trajectory equation, with some participants questioning the assumptions made about the puck's motion. There is acknowledgment of a provided solution from a professor, but confusion remains regarding the simplification of terms and the origin of specific values in the equation.

Contextual Notes

Participants note the importance of showing work to clarify reasoning and the challenge of addressing the problem without complete information. The discussion reflects on the constraints of the problem setup and the need for clarity in the equations used.

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Homework Statement


A rocket powered hockey puck on frictionless ice slides along the y-axis. The front of the rocket is tilted at an angle from the x-axis. The rocket motor ignities as the puck crosses the origin, exerting a Force(thrust) on the puck.
A. Find algebraic equation y(x) for the pucks trajectory
B. Thrust= 2N.
Mass of rocket/puck togther =1kg
Intial speed is 2.0m/s^2 along the y-axis. Make graph of f(x) at 45 deg. and -45 deg. fromx=0 to 20m

Homework Equations


y=y(nought)+vt+.5at^2
x=x(nought)+vt+.5at^2
a=F(net)/m

The Attempt at a Solution


A. I get: y(x)=V*(sqrt(2x/a*cos))+xtan
But it is incorrect.
B. at -45 deg, y(8.5) is suppose to equal 0 but my equation doesn't follow this.
 
Last edited:
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Its always difficult to answer these questions without the work shown. What would help is to show the x and y accelerations as a function of angle. In one case the puck will try to fly opposed by gravity, in the -45, the rocket thrust will add to gravity. In both cases the x acceleration is some fraction of the rocket force.
 
actually, the puck does not leave the ground; the puck is just in a x/y plane on ice taking on a projectile motion.
For X: x= .5a(x)cost*t^2, a(x) is acceleration in the x direction.
Y: y=vt+.5a(y)sin*t^2, a(y) is the acceleration in the y direction.
I solved for t in my x equation and got t=sqrt(x/a*cos)
Substituting t into Y I get y(x)=V*(sqrt(2x/a*cos))+xtan.
a=2m/s^2 so the equation can simplify to
y(x)=V*(sqrt(x/cos))+xtan.
 
Last edited:
lets call a(x)=a1. So x=.5*a1*cos(theta)*t^2 as initial Vx=0;

then t=sqrt(2x/(a1*cos(theta))) as in somewhere the .5 got dropped?
 
ok, my prof. gave us the solution to this problem today. I did get the generic equation right but still the wrong value. He said:
y=sqrt((2*1kg*x)\(2N*cos45))+xtan45; this simplyfies to :
y=2.38*sqrt(x)+x, but how? where does this 2.38 come from? xtan45=1x but what happens to the other terms?
 
Scratch that last note, I got it
 

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