# A rocket-powered hockey puck

1. Mar 12, 2009

### tatertot560

1. The problem statement, all variables and given/known data
A rocket-powered hockey puck has a thrust of 2.40 N and a total mass of 2.40 kg. It is released from rest on a frictionless table, 3.80 m from the edge of a 2.30 m drop. The front of the rocket is pointed directly toward the edge.

How far does the puck land from the base of the table?

3. The attempt at a solution
in solving the problem, this is what i got:

2.40 N = 0+1/2(9.8)t^2
4.80=9.8*t^2
sqrt(0.49)=t
t=0.70 s

F=ma
240N=240kg*a
a=1 m/s^2

v_x^2-v_o^2=2aS
v_x^2=2(1 m/s^2)(3.80 m)
sqrt(2*1*3.80)
v_x=2.76 m/sec

x=(2.76 m/sec)(0.70 sec)
x= 1.93 m

I plugged that in as the answer but i was told that it is wrong. Can someone please check my work? i'd really appreciate it :D

2. Mar 12, 2009

### lanedance

what is this this equation? Force does not equal 1/2. a.t^2 looks like a distance

do you mean 2.3m for the puck to fall from the table?
horizontal acceleration of the puck - looks good
velocity at edge of table due to rocket...
does the puck keep accelerating horizontally after it leaves the table?

3. Mar 12, 2009

### tatertot560

yeah i made a mistake with the number. i got 0.685 sec for the time when i plugged 230 m in instead of 2.40 N.

for that answer, i got 1.90 meters but i was still wrong.
as for accelerating horizontally, wouldn't it be accelerating vertically instead of horizontally?

4. Mar 12, 2009

### tatertot560

mistake: i didn't use the whole equation to figure out the horizontal acceleration.
i should have used this equation:
x=v_f*T +1/2 at^2

thanks for all the help. i really appreciate it :D