Minimum Horizontal Velocity Calculation for Stunt Person Jump

What is the minimum horizontal velocity required for a stunt person to jump between buildings?

Given the distance between the 2 buildings is 3 meters and the drop is 5 meters below, how can we calculate the minimum horizontal velocity needed for this jump?

Minimum Horizontal Velocity Calculation for Stunt Person Jump

The minimum horizontal velocity required for a stunt person to jump between buildings, 3 meters apart and 5 meters below, can be found using kinematic equations.

Explanation: To calculate the minimum horizontal velocity that a stunt person must have to make a successful jump from one building to another, we can ignore air resistance and focus on the purely horizontal motion. Since the buildings are 3 meters apart and the stunt person drops 5 meters, we can use the kinematic equations for uniformly accelerated motion.

The time (t) taken to fall 5 meters can be found using the equation for vertical motion (under the influence of gravity):

Distance (d): 5 meters

Acceleration due to gravity (g): 9.81 m/s²

Initial vertical velocity (u): 0 m/s (since the stunt person jumps horizontally)

We can use the following equation for vertical motion:

d = u*t + 0.5*g*t²

Plugging the values, we get:

5 = 0 + 0.5*9.81*t²

From this, we can solve for (t), the time in seconds, which comes out to be (t) < 1 second. Then, using (t), we can find the required horizontal velocity (v) to cover the 3-meter gap using the formula:

Horizontal distance (s) = horizontal velocity (v) * time (t)

3 = v*t

By dividing the horizontal distance by the time calculated earlier, we find the minimum horizontal velocity.