The mens world record for the shot put, 23.12 meters. if the shot was launched from 6 feet above the ground at an initial angle of 42 degrees, what was the initial speed?

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To determine the initial speed of the shot put, we can use the equations of projectile motion. The horizontal and vertical components of the shot put's motion can be analyzed separately.

First, let's consider the vertical motion. The shot put was launched at an initial angle of 42 degrees from a height of 6 feet (which is approximately 1.83 meters). The vertical displacement can be calculated as the difference between the initial and final heights:

Vertical displacement (Δy) = 23.12 meters - 1.83 meters = 21.29 meters

Using the formula for vertical displacement in projectile motion, we can find the initial vertical velocity (v₀y) of the shot put:

v₀y² = u² + 2as

Here, u represents the initial vertical velocity (which is what we are trying to find), a is the acceleration due to gravity (-9.8 m/s²), and s is the vertical displacement. Since the shot put was launched vertically upward, its final velocity (v) at the topmost point will be zero.

So, the equation becomes:

0² = v₀y² + 2(-9.8 m/s²)(21.29 m)

Simplifying the equation, we get:

(-9.8 m/s²)(21.29 m) = v₀y²

-207.722 m²/s² = v₀y²

Taking the square root of both sides, we can find the magnitude of initial vertical velocity:

v₀y ≈ 14.4 m/s

Now, let's consider the horizontal motion. The horizontal displacement (Δx) of the shot put is not provided, but it doesn't affect the calculation of the initial speed. The initial horizontal velocity (v₀x) remains constant throughout the motion.

Using the formula for horizontal displacement in projectile motion:

Δx = v₀x * t

Since the shot put was launched at an angle of 42 degrees, its initial horizontal velocity can be found using trigonometry:

cos(42°) = v₀x / v₀

Simplifying the equation, we get:

v₀x ≈ v₀ * 0.7431

To get the initial speed (v₀) of the shot put, we can divide the initial vertical velocity (v₀y) by the cosine of the launch angle:

v₀ = v₀y / cos(42°)

Substituting the values we calculated earlier:

v₀ ≈ 14.4 m/s / cos(42°)

v₀ ≈ 14.4 m/s / 0.7431

v₀ ≈ 19.38 m/s

Therefore, the initial speed of the shot put when launched from 6 feet above the ground at an angle of 42 degrees is approximately 19.38 m/s.