What initial velocity is required to achieve a high jump of 1.52 meters from a standing position?

To determine the initial velocity needed to achieve a high jump of 1.52 meters, we can use the principles of physics, specifically the equations of motion under the influence of gravity.

When a person jumps, they convert their initial kinetic energy into gravitational potential energy at the peak of the jump. The relationship can be described as follows:

The potential energy (PE) at the peak of the jump can be expressed as:

PE = m * g * h

Where:

  • m = mass (which we will not need since it cancels out)
  • g = acceleration due to gravity (approximately 9.81 m/s²)
  • h = maximum height (1.52 meters)

The initial kinetic energy (KE) at the takeoff can be represented as:

KE = (1/2) * m * v²

Where v is the initial velocity. Setting the kinetic energy equal to the potential energy at the maximum height gives us:

(1/2) * m * v² = m * g * h

We can cancel out the mass (m) from both sides, leading to:

(1/2) * v² = g * h

Now, we can solve for v:

v² = 2 * g * h
v = sqrt(2 * g * h)

Substituting in the known values:

g = 9.81 m/s²
h = 1.52 m
v = sqrt(2 * 9.81 * 1.52)
v = sqrt(29.87)
v ≈ 5.47 m/s

Therefore, the initial velocity required to achieve a high jump of 1.52 meters from a standing position is approximately 5.47 meters per second.

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