Projectile Motion Problem -Diving off cliff

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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 .

Diagram of a diver running off a cliff with horizontal launch velocity v0 to the right; cliff height labeled 9.00 m, horizontal distance to a ledge labeled 1.75 m, with water below the ledge.


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
 
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.
 
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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: