Projectile Motion Problem -Diving off cliff

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Homework Help Overview

The problem involves a swimmer diving off a cliff, analyzing the projectile motion in both horizontal and vertical directions. The swimmer's weight is given, but its relevance to the motion is questioned.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the use of motion charts for both x and y components, questioning the necessity of the swimmer's weight. They explore equations related to time of flight and horizontal distance.

Discussion Status

Some participants have made progress in calculating the time of flight and the initial horizontal velocity. There is an ongoing exploration of the equations of motion, with no explicit consensus reached on the relevance of the swimmer's weight.

Contextual Notes

The original poster expresses uncertainty about how to utilize the given weight of the swimmer, indicating a potential gap in understanding the relationship between weight and projectile motion.

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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