# Projectile in Motion

## 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 doesnt follow this.

Last edited:

Anyone?...

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