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As @Peter Mitrano suggests, the force is applied over only 1 simulation time step, resulting in a small impulse and therefore a very small change in velocity.

When I run Gazebo simulator, the default maxtime step is 0.001 (1 ms)

F = ma
v = at
v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then
v(t) = v(t=0) + (F*t)/m

Given:
v(t=0) = 0
F = 100 N (F > Fgravity)
t = 0.001 s (1 simulation time step)
m = 5 kg

v(0.001) = 0 + (1000.001)/5
v(0.001) = 0.02 m/s *or 2 cm/s

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

As @Peter Mitrano suggests, the force is applied over only 1 simulation time step, resulting in a small impulse and therefore a very small change in velocity.

When I run Gazebo simulator, the default maxtime step is 0.001 (1 ms)

F = ma
v = at
v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then
v(t) = v(t=0) + (F*t)/m

Given:
v(t=0) = 0
F = 100 N (F > Fgravity)
t = 0.001 s (1 simulation time step)
m = 5 kg

v(0.001) = 0 + (1000.001)/5
v(0.001) = 0.02 m/s *or 2 cm/s

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

As @Peter Mitrano suggests, the force is applied over only 1 simulation time step, resulting in a small impulse and therefore a very small change in velocity.

When I run Gazebo simulator, the default maxtime step is 0.001 (1 ms)

F = ma
v = at
v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then
v(t) = v(t=0) + (F*t)/m

Given:
v(t=0) = 0
F = 100 N (F > Fgravity)
t = 0.001 s (1 simulation time step)
m = 5 kg

v(0.001) = 0 + (1000.001)/5 (100*0.001)/5
v(0.001) = 0.02 m/s *or or 2 cm/s

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

As @Peter Mitrano suggests, the force is applied over only 1 simulation time step, resulting in a small impulse and therefore a very small change in velocity.

When I run Gazebo simulator, the default maxtime step is 0.001 (1 ms)

F = ma
v = at
v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then
v(t) = v(t=0) + (F*t)/m

Given:
v(t=0) = 0
F = 100 N (F > Fgravity)
t = 0.001 s (1 simulation time step)
m = 5 kg

v(0.001) = 0 + (100*0.001)/5
v(0.001) = 0.02 m/s or 2 cm/s

Throwing in the effects of gravity, we can determine (roughly) that the 5 kg sphere in your example should - in theory - reach a maximum vertical (+z) height of 0.01 m (1 cm) and return to a height of 0 m after 0.02 s (20 ms).

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

As @Peter Mitrano suggests, the force is applied over only 1 simulation time step, resulting in a small impulse and therefore a very small change in velocity.

When I run Gazebo simulator, the default maxtime step is 0.001 (1 ms)

F = ma
v = at
v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then
v(t) = v(t=0) + (F*t)/m

Given:
v(t=0) = 0
F = 100 N (F > Fgravity)
t = 0.001 s (1 simulation time step)
m = 5 kg

v(0.001) = 0 + (100*0.001)/5
v(0.001) = 0.02 m/s or 2 cm/s

Throwing in the effects of gravity, we can determine (roughly) that the 5 kg sphere in your example should - in theory - reach a maximum vertical (+z) height of 0.01 m (1 cm) and return to a height of 0 m after 0.02 s (20 ms).

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

As @Peter Mitrano suggests, the force is applied over only 1 simulation time step, resulting in a small impulse and therefore a very small change in velocity.

When I run Gazebo simulator, the default maxtime step is 0.001 (1 ms)

F = ma
v = at
v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then
v(t) = v(t=0) + (F*t)/m

Given:
v(t=0) = 0
F = 100 N (F > Fgravity)
t = 0.001 s (1 simulation time step)
m = 5 kg

v(0.001) = 0 + (100*0.001)/5
v(0.001) = 0.02 m/s or 2 cm/s

Throwing in the effects of gravity, we can determine (roughly) that the 5 kg sphere in your example should - in theory - reach a maximum vertical (+z) height of 0.01 m (1 cm) and return to a height of 0 m after 0.02 s (20 ms).

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

As @Peter Mitrano suggests, the force is applied over only 1 simulation time step, resulting in a small impulse and therefore a very small change in velocity.

When I run Gazebo simulator, the default maxtime step is 0.001 (1 ms)

F = mm * a
v = aa * t
v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then
v(t) = v(t=0) + (F*t)/m(F * t)/m

Given:
v(t=0) = 0
F = 100 N (F > Fgravity)
t = 0.001 s (1 simulation time step)
m = 5 kg

v(0.001) = 0 + (100*0.001)/5 (100 * 0.001)/5
v(0.001) = 0.02 m/s or 2 cm/s

Throwing in the effects of gravity, we can determine (roughly) that the 5 kg sphere in your example should - in theory - reach a maximum vertical (+z) height of 0.01 m (1 cm) and return to a height of 0 m after 0.02 s (20 ms).

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection

As @Peter Mitrano suggests, the force is applied over only 1 simulation time step, resulting in a small impulse and therefore a very small change in velocity.

When I run Gazebo simulator, the default maxtime step max time (step) size is 0.001 (1 ms)

F = m * a
v = a * t
v(t) = v(t=0) + 1/m * integral(F(t), 0, t)

If we assume a constant F across the time step, then
v(t) = v(t=0) + (F * t)/m

Given:
v(t=0) = 0
F = 100 N (F > Fgravity)
t = 0.001 s (1 simulation time step)
m = 5 kg

v(0.001) = 0 + (100 * 0.001)/5
v(0.001) = 0.02 m/s or 2 cm/s

Throwing in the effects of gravity, we can determine (roughly) that the 5 kg sphere in your example should - in theory - reach a maximum vertical (+z) height of 0.01 m (1 cm) and return to a height of 0 m after 0.02 s (20 ms).

Physics: http://www.physicsclassroom.com/class/momentum/Lesson-1/Momentum-and-Impulse-Connection