Velocity to jump building to building

[bmarvs04]

Homework Statement

"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? "

Homework 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

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

 

[LowlyPion]

Homework Statement

"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? "

Homework 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

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

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?

 

[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

 

[LowlyPion]

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

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.)

 

[bmarvs04]

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