Projecting Stunt Man: A Physics Adventure

How long was the stunt man in the air for?

What was his horizontal velocity when he came out of the cannon?

The stunt man was in the air for 2 seconds and his horizontal velocity when he came out of the cannon was 75 m/s.

Imagine being at a circus and witnessing a fascinating stunt involving a man climbing up 19.6 meters into a cannon and getting fired horizontally, landing 150 meters away. To calculate the duration of his flight and his velocity upon exiting the cannon, we delve into the realm of physics.

Acceleration, in physics, is the rate at which an object's velocity changes over time. It is a vector quantity, meaning it has both magnitude and direction. The direction of acceleration depends on the net force acting on an object. In this scenario, the stunt man experiences a downward acceleration due to gravity as he is fired horizontally.

Using the displacement formula for free falling objects, we can determine the time of flight. The vertical displacement 'h' is 19.6 meters, acceleration 'a' due to gravity is -9.8 m/s^2, and the initial vertical speed 'u' is 0 since he is launched horizontally. Plugging the values into the formula 'h = ut + 1/2 a*t^2', we find t = sqrt(2*h/a) = sqrt(2*19.6 / 9.8) = 2 seconds.

To calculate the horizontal velocity, we use the formula v = d/t, where the horizontal distance 'd' is 150 meters and the time 't' is 2 seconds, as calculated earlier. Thus, the horizontal velocity is v = 150/2 = 75 m/s.

Therefore, the stunt man spent 2 seconds in the air and had a horizontal velocity of 75 m/s when he emerged from the cannon, creating a thrilling physics spectacle at the circus.