Velocity to jump building to building

In summary, two individuals are discussing a physics problem involving a movie stuntman running off a building and landing on another building 1.6 m lower. The problem asks for the required speed for the stuntman to successfully make the jump, and the individual attempts a solution using the equation Vf^2 = V0^2 + 2a (Xf-X0). However, they realize that the hypotenuse is not the appropriate approach and instead consider the time it takes for the stuntman to fall 1.6 m and use that to calculate the necessary horizontal velocity. The final answer is determined to be 1.876 m/s.
  • #1
bmarvs04
12
0

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
 
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  • #2
bmarvs04 said:

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?
 
  • #3
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
bmarvs04 said:
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.)
 
  • #5
Thanks a bunch! That makes perfect sense to me now. You've been a great help!
 

1. What is the average velocity required to jump from one building to another?

The average velocity required to successfully jump from one building to another will vary depending on the distance between the buildings and the height of the buildings. However, it is estimated that a velocity of at least 20 miles per hour is needed to clear a gap of 20 feet between buildings.

2. How does air resistance affect the velocity needed for a building-to-building jump?

Air resistance, also known as drag, can significantly impact the velocity needed for a building-to-building jump. As the velocity increases, so does the amount of air resistance. This means that a higher velocity is required to overcome the drag and successfully complete the jump.

3. Is it possible to calculate the exact velocity needed for a specific building-to-building jump?

Yes, it is possible to calculate the exact velocity needed for a specific building-to-building jump using the principles of physics. Factors such as the distance between buildings, height of the buildings, and air resistance can be taken into account to determine the precise velocity needed for a successful jump.

4. Can a person increase their jumping velocity through training?

Yes, a person can increase their jumping velocity through training. Techniques such as plyometrics, resistance training, and proper form can help improve a person's jumping ability. However, it is important to note that attempting dangerous jumps between buildings is not recommended and can result in serious injury or death.

5. Are there any safety precautions that should be taken when attempting a building-to-building jump?

Yes, there are several safety precautions that should be taken when attempting a building-to-building jump. These include wearing appropriate safety gear, ensuring the jump is legal and authorized, and practicing proper technique. It is also important to have a trained professional present and to never attempt a jump that is beyond one's abilities.

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