# Minimum Horizontal Velocity for a Stunt Person Jump

## How can we determine the minimum horizontal velocity needed for a stunt person to leap from one building to another?

In a movie production, a stunt person must leap from a balcony of one building to a balcony 3.0 m lower on another building. If the buildings are 2.0 m apart, what is the minimum horizontal velocity the stunt person must have to accomplish the jump?

## Minimum Horizontal Velocity Calculation

The minimum horizontal velocity required for the stunt person to successfully make the jump is 2.6 m/s.

## Explanation of Minimum Horizontal Velocity Calculation

To calculate the minimum horizontal velocity required for a stunt person to successfully leap from one building's balcony to another that is 3.0 m lower, we can use projectile motion equations. The equation for the time it takes an object to fall a certain vertical distance when the initial vertical velocity is zero is t = √(2y/g), where g is the acceleration due to gravity and is approximately 9.81 m/s^2. When the stunt person jumps, they will have to cover the horizontal distance of 2.0 m during the same time interval.

The time taken to fall 3.0 m can be calculated as:

t = √(2y/g) = √(2*3.0 m / 9.81 m/s^2) = √(0.612 m/s^2) = 0.782 s

Therefore, the minimum horizontal velocity vx must be:

vx = d/t = 2.0 m / 0.782 s = 2.56 m/s

The stunt person must therefore have a minimum horizontal velocity of at least 2.56 m/s to make the jump.