What is the Minimum Speed for a Diver to Clear a Cliff Ledge?

AI Thread Summary
The discussion revolves around calculating the minimum speed a diver needs to clear a cliff ledge that is 1.50 meters wide and 9.50 meters below the top. The initial approach involved using kinematic equations to determine the time of fall and horizontal distance, but the calculations were incorrect. Participants suggested leaving variables in the equations to derive a formula for initial velocity based on given distances. The conversion of the final speed from meters per second to miles per hour was also discussed, emphasizing that the initial speed in both units should yield the same result. The thread highlights the importance of correctly applying physics principles to solve the problem.
Ecterine
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A daring swimmer dives off a cliff with a running horizontal leap, as shown in the figure.

http://smg.photobucket.com/albums/v113/apotheothenai/?action=view&current=1011380A.jpg

Part A) What must the diver's minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.50m wide and 9.50m below the top of the cliff?


I tried to use y = y0 + vy0*t - 1/2 * g * t^2 and then x = vx0 * t

y = y0 + vy0*t - 1/2 * g * t^2
0 = 9.50 + 0*t - 1/2 * 9.8 * t^2 (plugged stuff in)
0 = 9.50 - 4.9 *t^2 (simplified)
-t^2 = 4.6 (square root of both sides)
t = 2.14

Then,
x = vx0 * t
x = 1.50 * 2.14
x = 3.14

It didn't work... :/
I still don't know what I'm doing in this class.


Part B) What must the diver's initial speed be in miles per hour? - I know this is a dumb question, but will this be the same as the minimum speed, except in MPH?
 
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for part a it seems you had it right but what is 1.50 in the equation x=vx0 *t ? shouldn't it be meters and thus x=1.50? vx0 is what you're looking for.

Also it might be a bit useful for you to leave the variables and find an equation that will give you the initial velocity required if all you are given is X and Y distance.
You have it done already just you replaced the variables by the known data right away.

and for part b yes, when you get part a just convert that to mph.
 
Last edited:
So instead...

1.50 = vx0 * 2.14 (divide both sides by 2.14)
.700

It didn't work... :/

I'm really not good at this
 
Ecterine said:
A daring swimmer dives off a cliff with a running horizontal leap, as shown in the figure.

http://smg.photobucket.com/albums/v113/apotheothenai/?action=view&current=1011380A.jpg

Part A) What must the diver's minimum speed be just as she leaves the top of the cliff so that she will miss the ledge at the bottom, which is 1.50m wide and 9.50m below the top of the cliff?


I tried to use y = y0 + vy0*t - 1/2 * g * t^2 and then x = vx0 * t

y = y0 + vy0*t - 1/2 * g * t^2
0 = 9.50 + 0*t - 1/2 * 9.8 * t^2 (plugged stuff in)
0 = 9.50 - 4.9 *t^2 (simplified)
-t^2 = 4.6 (square root of both sides)

what happens here? this step isn't right.
 
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