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
Messages
13
Reaction score
0
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?
 
Physics news on Phys.org
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.
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top