Minimum Speed for a Horizontally Projected Diver to Clear a Ledge

AI Thread Summary
To determine the minimum speed a 510N swimmer must achieve to clear a 1.75m ledge while diving from a height of 9.00m, projectile motion principles are applied. The key equations used are s=1/2gt^2 and v=s/t, emphasizing that the swimmer's weight does not affect the calculations since all projectiles experience the same acceleration due to gravity. The inclusion of the swimmer's weight appears to be a test of understanding that gravitational acceleration is independent of mass. The discussion highlights the common confusion among students regarding the relevance of weight in such problems. Ultimately, the focus remains on mastering the basics of projectile motion.
dregozo

Homework Statement


A daring 510N swimmer dives off a cliff with a horizontal leap. What must his minimum speed be just as he leaves the top of the cliff so that he will miss the ledge at the bottom, which is 1.75m wide and 9.00m below the top of the cliff?

Homework Equations


s=1/2gt^2 ; v=s/t

The Attempt at a Solution


I used the above equations in that order, and arrived at an answer. My issue is that I did not use the 510N at all! Where does the diver's weight come in?
 
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dregozo said:
Where does the diver's weight come in?
It doesn't! This is a projectile motion problem and all projectiles have the same acceleration, independent of mass.
 
My exact reasoning.. so why did the examiner include the 510N?? To test students if they know that g is independent of mass?
 
dregozo said:
so why did the examiner include the 510N?? To test students if they know that g is independent of mass?
Sure, why not? That's one good reason to include it -- to see how confident you are with the basics.

Is this just a part of a multi-part question?
 
No, it's the entire question. Nothing more was asked. Strange, right?
 
It happens often. Keeps you on your toes! (And drives students nuts!) :smile:
 
I wish I weighed 510N :-(
 
CWatters said:
I wish I weighed 510N :-(
LOL. (Me too!)
 

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