# Velocity to jump building to building

1. Sep 4, 2008

### bmarvs04

1. The problem statement, all variables and given/known data

"In a chase scene, a movie stuntman is supposed to run right off the flat roof of one city building and land on another roof 1.6 m lower."

"If the gap between the buildings is 4.5 m wide, how fast must he run? "

2. Relevant equations

Vf^2 = V0^2 + 2a (Xf-X0)

This might not be the right equation, but it's the one I've been trying to use

3. The attempt at a solution

Vf^2 = 2(9.8)(4.776)

4.776 is the hypotenuse of the 1.6m and 4.5m distances

2. Sep 4, 2008

### LowlyPion

Welcome to PF.

Unfortunately the hypotenuse would not be the right approach.

Maybe figure how much time it would take Sammy Stunt guy to fall 1.6 m?

Then you would know how much time he has to travel the 4.5 m at constant velocity?

3. Sep 4, 2008

### bmarvs04

Ok thanks,

So in order to solve for the vertical velocity, I used the equation Vf^2 = 2(9.8)(1.6) = 5.6m

Then do I use trig to solve for horizontal velocity? In other words, Horizontal Velocity = cos(70.4268)*5.6 where 70.4268 is the downward angle of the triangle formed by 1.6m and 4.5m.

I got 1.876 m/s as my answer, does this sound right?

Thanks

4. Sep 5, 2008

### LowlyPion

No. Not exactly. The presumption is the runner is running horizontally. So the runner only has to worry about going 4.5 meters before he drops 1.6 meters.

He drops 1.6 meters in how many seconds? x = 1/2 a*t^2

That equals 1.6 * 2 / 9.8 = t^2

Then divide that time into 4.5 m.

That way the stunt guy manages to go 4.5 m before he drops more than 1.6 m. (Because if he drops more than 1.6 m he drops a lot more than that.)

5. Sep 5, 2008

### bmarvs04

Thanks a bunch! That makes perfect sense to me now. You've been a great help!