Projectile Motion in Circus: Calculating Stunt Man's Time in the Air

Understanding Projectile Motion in Circus Stunt

When the stunt man climbs up 29.4 meters into the cannon and gets fired horizontally out of it with a speed of 57.1 m/s, we can apply the physics concept of projectile motion to calculate how long he was in the air. In this scenario, we are dealing with horizontal projectile motion, where the vertical motion is determined solely by gravity.

The key equation we use to solve for the time the stunt man was in the air is d = 0.5at², where d represents the distance fallen (29.4 meters), a is the acceleration due to gravity (9.8 m/s²), and t is the time we are solving for.

Given that the initial vertical velocity (v) is 0 m/s since the stunt man was fired horizontally, the equation simplifies to d = 0.5at². Plugging in the known values and solving for t, we get:

t = sqrt[(2*d)/a] = sqrt[(2*29.4 m)/9.8 m/s²] = sqrt[6] ≈ 2.45 s.

Therefore, we conclude that the stunt man was in the air for approximately 2.45 seconds after being fired out of the cannon at a speed of 57.1 m/s from a height of 29.4 meters.