The best leaper in the animal kingdom is the puma, which can jump to a height of 12.4 ft when leaving the ground at an angle of 38°. With what speed, in SI units, must the animal leave the ground to reach that height?

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To determine the speed at which the puma must leave the ground to reach a vertical height of 12.4 ft, we can use basic kinematic equations and principles.

First, let's convert the given height from feet to meters, as the SI units are in meters. Since 1 ft is approximately equal to 0.3048 m, the height of 12.4 ft is equal to 12.4 ft * 0.3048 m/ft = 3.77952 m.

Now, we can break down the motion of the puma into horizontal and vertical components. The 12.4 ft height corresponds to the vertical component, and the given angle of 38° tells us the launch angle of the puma.

The vertical component of the puma's initial velocity can be determined using the formula:
V₀y = V₀ * sin(θ)

Where:
V₀y represents the vertical component of the initial velocity,
V₀ represents the initial velocity, and
θ represents the angle of launch.

Rearranging the formula, we have:
V₀ = V₀y / sin(θ)

Substituting the known values:
V₀ = 3.77952 m / sin(38°)

Next, we need to convert the angle from degrees to radians since trigonometric functions in most programming languages and calculators work with radians. The conversion is π radians = 180°, so:

θ_radians = 38° * π/180°

Substituting the angle in radians:
V₀ = 3.77952 m / sin(38° * π/180°)

Evaluating this expression will give us the speed at which the puma must leave the ground to reach a height of 12.4 ft.