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

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Homework Help Overview

The discussion revolves around a physics problem involving projectile motion, specifically focusing on the minimum speed required for a diver to clear a cliff ledge. The diver must leap horizontally from a height of 9.50 meters to avoid a ledge that is 1.50 meters wide.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants discuss the equations of motion for vertical and horizontal components, questioning the application of variables and the setup of the problem. There are attempts to solve for the diver's initial speed using both vertical and horizontal motion equations.

Discussion Status

Some participants are providing feedback on the original poster's calculations, suggesting that the values used in the equations may need to be reconsidered. There is a recognition of confusion regarding the steps taken in the calculations, particularly in determining the correct initial velocity.

Contextual Notes

Participants express uncertainty about the calculations and the interpretation of the problem's parameters. There is an acknowledgment of the need to clarify the relationship between the horizontal distance and the initial speed.

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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