# Another aggravating kinematics problem - regarding stones

1. Sep 21, 2006

### OVB

A stone is DROPPED off a 144 foot cliff. A second later, another stone is THROWN off that cliff. The two hit the water at the same exact instant. What is the initial velocity of the second stone?

The answer is 40 ft/second according to my book, but for some reason, I never get that.

I set 4.9t^2 = v*(t-1) + 4.9(t-1)^2
I also set 144 = 4.9t^2 to get t = 5.421.

I plug this back into the first equation, and my answer comes out to 10.908 ft/ sec!

What am I doing wrong here?

2. Sep 21, 2006

### QuantumKing

youve got the gravitational acceleration wrong...im pretty sure its not 4.9f/s^2...if its 9.8m/s^2

3. Sep 21, 2006

### QuantumKing

its approximately 32.2 f/s^2

4. Sep 21, 2006

Interesting, I get the same answer. From the equation of displacement for the first rock we get t = 5.42 seconds. The time of 'flight' of the second stone must be 4.42 seconds. So, x2(t)=v0*4.42+0.5*9.81*4.42^2=144 implies v0=10.9 ft/second.

5. Sep 21, 2006

### OVB

OHHH darn I forgot to convert the units. Silly me :P

6. Sep 22, 2006