Newton's second law and kinematics allow to find the result for the velocity at the point of reaching the child's net is:
v = 23.7 m / s
Newton's second law states that the net force is proportional to the mass and acceleration of the body.
∑ F = m a
Where F is the force, m the mass and the acceleration of the body.
In the attachment we can see a free body diagram of the system, where the positive direction is up.
F -W = m a
Weight is
W = m g
[tex]a = \frac{F}{m} - g[/tex]
Let's calculate
[tex]a= \frac{30}{70} - 9.8[/tex]
a = -9.37 m / s²
Negative indicates downward direction.
Kinematics studies the movement of the body, let's look for speed.
v² = v₀² - 2 a (y-y₀)
Where v is the speed, v₀ is the initial speed, a tha acceleration ang (y-y₀) is the eight.
As the child is released, his initial velocity is zero, upon reaching the floor his height is zero and his initial height is I = 30 m.
[tex]v= \sqrt{2 a y_o}[/tex]
Let's calculate
v = [tex]\sqrt{2 \ 9.37 \ 30}[/tex]
v = 23.7 m / s
In conclusion using Newton's second law and kinematics we can find the velocity at the point where the child reaches the net is:
v = 23.7 m / s
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The velocity of the child before hitting the ground is 22.596 m/s.
The given parameters;
The weight of the child acting downwards is calculated as follows;
W = mg
W = 70 x 9.8
W = 686 N
The net force on the child is calculated as follows;
∑Fₙ = 686 N - 90 N
∑Fₙ = 596 N
The acceleration of the child is calculated as follows;
F = ma
[tex]a = \frac{F}{m} \\\\a = \frac{596}{70} \\\\a = 8.51 \ m/s^2[/tex]
The velocity of the child before hitting the ground is calculated as;
v² = u² + 2as
v² = 0 + (2 x 8.51 x 30)
v² = 510.6
v = √510.6
v = 22.596 m/s
Thus, the velocity of the child before hitting the ground is 22.596 m/s.
Learn more here:https://brainly.com/question/9240794
