Ballistic Motion using Newton's Laws

[JeYo]

A rocket-powered hockey puck has a thrust of 1.20 and a total mass of 1.50 . It is released from rest on a frictionless table, 3.20 from the edge of a 3.60 drop. The front of the rocket is pointed directly toward the edge. How far does the puck land from the base of the table?



Okay, so I found the accleration of the puck in the x-direction to be 0.8m/s/s and the difference in time between the moment it is at the end of the table to be 0.857s. But past this I have been unable to find initial velocity or final velocity or anything that I could plug into a kinematics equation to help me find the final position of the puck, on the x-axis.

 

[Astronuc]

One should include units, but it appears one is using SI or mks.

It is released from rest on a frictionless table

and accelerates at 0.8 m/² over a distance of 3.2 m.

http://hyperphysics.phy-astr.gsu.edu/hbase/mot.html#mot5

The puck then falls 3.6 m under the influence of gravity with no initial vertical velocity, but it has some horizontal velocity and perhaps horizontal acceleration(?).

http://hyperphysics.phy-astr.gsu.edu/hbase/traj#tra11

ref - http://hyperphysics.phy-astr.gsu.edu/hbase/traj.html

 

[ogb p]

I would approach this problem by using Newton's Laws of Motion to analyze the motion of the puck. Since the puck is released from rest on a frictionless table, we can assume that there is no external force acting on it in the horizontal direction. Therefore, the only force acting on the puck is the thrust from the rocket, which is equal to its mass multiplied by its acceleration.

Using Newton's Second Law (F=ma), we can calculate the acceleration of the puck to be 0.8 m/s^2. We can then use this acceleration and the distance from the edge of the table (3.20 m) to calculate the time it takes for the puck to reach the edge of the table using the equation d=0.5at^2. This gives us a time of 2.27 seconds.

Next, we can use the equation v=at to find the initial velocity of the puck, which is 1.82 m/s. This is also the final velocity of the puck when it reaches the edge of the table.

To find the final position of the puck, we can use the equation x=vt to calculate the distance traveled in the x-direction. Plugging in the values of initial velocity and time, we get a distance of 4.14 m. This means that the puck will land 4.14 m from the base of the table.

In conclusion, by using Newton's Laws of Motion and kinematics equations, we can determine that the puck will land 4.14 m from the base of the table when released from rest with a thrust of 1.20 N and a total mass of 1.50 kg.