Calculating Shot Put Release Speed | Projectile Motion Homework Problem

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The discussion centers on calculating the release speed of a shot put thrown at a 40-degree angle, referencing Randy Barnes' world record throw. The shot put weighs 7.30 kg and is released from a height of 2.00 m. Participants express confusion over the phrase "40 degrees over a straight left leg," clarifying that it likely refers to an angle above the horizontal axis. Key equations for projectile motion are provided to aid in the calculation. The focus remains on determining the initial speed required to achieve the record distance.
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Homework Statement


In the shot put, a standard track-and-field event, a 7.30 kg object (the shot) is thrown by releasing it at approximately 40.0 degrees over a straight left leg. The world record for distance, set by Randy Barnes in 1990, is 23.11 m. Assuming that Barnes released the shot put at 40.0 degrees from a height of 2.00 m above the ground, with what speed, in m/s, did he release it?


Homework Equations



x-xo=vox*t

y=yo+voy*t+1/2*a*t^(2)

The Attempt at a Solution



Well I was a bit confused with the wording on this sentence:
In the shot put, a standard track-and-field event, a 7.30 kg object (the shot) is thrown by releasing it at approximately 40.0 degrees over a straight left leg.

It says 40 degrees over a straight left leg. What does that actually mean? Thanks!
 
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I think 40 degrees over a straight left leg just means 40 degrees above the x-axis.
 
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