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A 40 kg child walks off the edge of a 10 M high dive platform and falls to the pool below. Hoss we fast is he moving right before he hits the water



Answer :

Intriguing456

Answer:

14 m/s

Explanation:

The child is falling due to constant gravitational acceleration.

This means we can use a kinematics equation to relate:

  • velocity
  • time
  • distance

First, we can list our knowns (givens):

  • distance = Δx = 10 m
  • mass = m = 40 kg
  • initial velocity = vᵢ = 0 m/s
  • gravitational acceleration = g = 9.8 m/s²

We are solving for:

  • final velocity

This means we can use the combined kinematics equation to relate the knowns and unknowns:

  • [tex]v_f^2 = v_i^2 + 2a\Delta x[/tex]

where:

  • [tex]v_f[/tex] = final velocity
  • [tex]v_i[/tex] = initial velocity
  • [tex]a[/tex] = acceleration
  • [tex]\Delta x[/tex] = change in position (distance traveled)

Plugging in the known values, we get:

[tex]v_f^2 = (0\text{ m/s})^2 + 2\!\left(9.8\text{ m/s}^2\right)\!(10\text{ m})[/tex]

Solving for [tex]v_f[/tex], we get:

[tex]v_f^2 = 196\text{ m}^2/\text{s}^2[/tex]

[tex]v_f = \sqrt{196\text{ m}^2/\text{s}^2}[/tex]

[tex]\boxed{v_f = 14\text{ m/s}}[/tex]

So, the child is moving 14 meters per second right before he hits the water.

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