To find the velocity with which the body leaves the ground, we can use Newton's second law of motion, which states that the force acting on an object is equal to the rate of change of its momentum. Mathematically, this can be expressed as:
F = Δp / Δt
Where: F = force acting on the object Δp = change in momentum of the object Δt = time interval during which the force is applied
The momentum of an object is given by the product of its mass (m) and velocity (v):
p = m * v
Rearranging the equation, we have:
Δp = m * Δv
Given that the mass (m) of the body is 5 kg and the force (F) acting on it is 600 N for a time interval (Δt) of 0.1 seconds, we can substitute these values into the equation:
600 N = 5 kg * Δv / 0.1 s
Rearranging the equation to solve for Δv:
Δv = (600 N * 0.1 s) / 5 kg
Δv = 12 m/s
Therefore, the change in velocity of the body is 12 m/s.
Since the body starts from rest (zero initial velocity) and is ejected vertically, the final velocity (v) with which the body leaves the ground is equal to the change in velocity (Δv):
v = Δv = 12 m/s
Thus, the body leaves the ground with a velocity of 12 m/s.