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.