A rocket-powered hockey puck has a thrust of 2.20 newtons and a total mass of 0.700 kilograms . It is released from rest on a frictionless table, 2.90 meters from the edge of a 2.80 meter drop. The front of the rocket is pointed directly toward the edge.

is there a question here?

how far does the puck land from the base of the table? and sorry i thought i included that

F*distance= 1/2 masspuck*velocity^2

solve for veloicty at the edge of the table.

How long does it drop?
h=1/2 g t^2 solve for t, time of drop.

How far horizontally?

distance= velocitiyhorizontal*timedrop

To solve this problem, we need to calculate the distance the rocket travels horizontally before it falls off the edge.

First, let's break down the problem and identify the key information given:

- Thrust of the rocket = 2.20 newtons
- Total mass of the rocket = 0.700 kilograms
- Distance from the edge of the table = 2.90 meters
- Length of the table = 2.80 meters

Now, let's analyze the forces acting on the rocket while it is on the table. There are two forces that affect the rocket's motion horizontally:

1. Thrust force (F_thrust) - This is the force provided by the rocket's engines, causing it to move forward.
2. Friction force (F_friction) - Since the table is assumed to be frictionless, there is no friction acting on the rocket.

Using Newton's second law of motion (F = ma), we can determine the acceleration of the rocket before it falls:

F_net = F_thrust - F_friction
F_net = F_thrust - 0 (since there is no friction)
F_net = F_thrust

F_net = ma
2.20 N = 0.700 kg * a

Now, we can solve for acceleration (a):

a = 2.20 N / 0.700 kg
a ≈ 3.14 m/s^2

Since the rocket is released from rest, we can use the kinematic equation to calculate the horizontal distance it travels before falling off:

d = v_0*t + (1/2)*a*t^2

Where:
- d = horizontal distance traveled
- v_0 = initial velocity (which is zero in this case since the rocket is initially at rest)
- a = acceleration
- t = time taken to fall

We need to find the time taken to fall. To do this, we'll use the kinematic equation for vertical motion:

d = v_0*t + (1/2)*g*t^2

Where:
- d = vertical distance fallen (2.80 meters)
- v_0 = initial velocity in the vertical direction (which is zero since the rocket is released from rest)
- g = acceleration due to gravity (9.8 m/s^2)
- t = time

Simplifying the equation:

2.80 m = (1/2)*9.8 m/s^2*t^2
5.60 m = 4.9 m/s^2*t^2
t^2 ≈ 5.60 m / 4.9 m/s^2
t^2 ≈ 1.14 s^2

Taking the square root of both sides:

t ≈ √(1.14 s^2)
t ≈ 1.07 s

Now that we have the time, we can calculate the horizontal distance (d):

d = v_0*t + (1/2)*a*t^2
d = 0*1.07 s + (1/2)*(3.14 m/s^2)*(1.07 s)^2
d ≈ 1.76 m

Therefore, the rocket will travel approximately 1.76 meters horizontally before falling off the edge.