Projectile Motion Problem -Diving off cliff

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SUMMARY

The discussion focuses on solving a projectile motion problem involving a swimmer diving off a cliff. The swimmer has a weight of 510N and performs a horizontal leap with a distance of 1.75m and a vertical drop of 9.00m. The calculations reveal that the time of flight is 1.35 seconds, leading to an initial horizontal velocity (VoX) of 1.30m/s. The use of kinematic equations, specifically ΔY = 1/2at² and ΔX = VoX * t, is essential for determining these values.

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Chandasouk
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Homework Statement


A daring 510N swimmer dives off a cliff with a running horizontal leap, as shown in the figure .

YF-03-39.jpg


I'm not sure how to begin, so I made a chart of the X and Y motions

X Motion Y Motion
--------- -----------------

VoX = ? Voy= 0m/s

/\X = 1.75m /\Y = 9.00m

a= -9.8m/s^2
t=?



I have no idea what to do with the 510N they give me or even if it is necessary
 
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I like your x and y motion chart - good start!
You will need to use some formulas to work out the initial velocity necessary to make it by the cliff. Ask yourself what kind of motion you have in the x direction and write down the appropriate equation(s). Same for the y direction.

Hot tip: you can make a perfect Δ here by copying one from https://www.physicsforums.com/blog.php?b=347
and pasting here.
 
Last edited by a moderator:
Hmm, I think I found the time.

ΔY = 1/2at^2

9.00m = 1/2(9.8m/s^2)t^2

9.00m
------ = t^2
4.9m/s^2

t=1.35 seconds


Knowing this, I could

ΔX = voxt

1.75m = vox(1.35sec)

vox = 1.30m/s ?
 
Chandasouk said:
Hmm, I think I found the time.

ΔY = 1/2at^2

9.00m = 1/2(9.8m/s^2)t^2

9.00m
------ = t^2
4.9m/s^2

t=1.35 seconds


Knowing this, I could

ΔX = voxt

1.75m = vox(1.35sec)

vox = 1.30m/s ?
Looks good :smile:
 

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