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

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SUMMARY

The minimum speed required for a diver to clear a cliff ledge that is 1.50 meters wide and 9.50 meters below the top of the cliff is calculated using projectile motion equations. The diver's vertical motion is governed by the equation y = y0 + vy0*t - 1/2 * g * t^2, where g is the acceleration due to gravity (9.8 m/s²). The diver must achieve a horizontal speed (vx0) of approximately 0.700 m/s to ensure she clears the ledge. This speed can be converted to miles per hour for practical understanding.

PREREQUISITES
  • Understanding of basic physics concepts, particularly projectile motion.
  • Familiarity with kinematic equations, specifically y = y0 + vy0*t - 1/2 * g * t^2.
  • Knowledge of unit conversion between meters per second and miles per hour.
  • Ability to manipulate algebraic equations to solve for unknown variables.
NEXT STEPS
  • Study the principles of projectile motion in physics.
  • Learn how to derive equations for initial velocity in projectile motion scenarios.
  • Practice converting units, specifically from meters per second to miles per hour.
  • Explore real-world applications of projectile motion, such as in sports or engineering.
USEFUL FOR

Students studying physics, educators teaching kinematics, and anyone interested in understanding the dynamics of projectile motion in real-life scenarios.

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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