# Projectile Motion of a swimmer

1. Sep 22, 2007

### Ecterine

A daring swimmer dives off a cliff with a running horizontal leap, as shown in the figure.

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?

2. Sep 22, 2007

### bob1182006

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: Sep 22, 2007
3. Sep 22, 2007

### Ecterine

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

It didn't work... :/

I'm really not good at this

4. Sep 22, 2007

### learningphysics

what happens here? this step isn't right.