The Physics Behind Circus Stunts

How long was the stunt man in the air for?

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

Answer:

The stunt man was in the air for 2 seconds and his initial horizontal velocity was 75 m/s.

When observing a daring stunt at a circus, it can be fascinating to analyze and understand the physics behind it. In this particular scenario, a stunt man climbed up 19.6 meters into a cannon and was fired horizontally, landing 150 meters away. The key questions revolve around the duration he spent in the air and his horizontal velocity upon exiting the cannon.

To determine how long the stunt man was in the air, we need to consider the principles of projectile motion and equations of linear motion under gravity. Since the stunt man was fired horizontally, he did not have any initial vertical velocity (Voy = 0). Calculating the time he spent in the air involves vertical motion. By using the equation h = 0.5gt^2, where h is the height of 19.6 meters and g is the acceleration due to gravity (9.8 m/s^2), we find that the time (t) is 2 seconds. Therefore, the stunt man was in the air for 2 seconds.

For his horizontal velocity upon exiting the cannon, we apply the fact that horizontally launched projectiles maintain a constant horizontal velocity (Vx). Using the formula Vx = x/t, with x being the horizontal displacement of 150 meters and t being the time of 2 seconds calculated earlier, we determine that the horizontal velocity was 75 m/s. Thus, the stunt man's initial horizontal velocity was 75 m/s.

Understanding projectile motion and the principles of physics involved in such stunts adds depth to the spectacle witnessed at the circus. It showcases how scientific concepts can be applied to real-world scenarios, enhancing our appreciation for the physics behind seemingly miraculous feats. To delve deeper into the topic of projectile motion, further exploration and study are recommended for a comprehensive grasp of this fascinating subject.