The discussion centers on calculating the maximum height and speed of a golf ball launched vertically using a homemade launcher. The time of flight is approximately 20 seconds, with 10 seconds ascending and 10 seconds descending. The theoretical calculation, ignoring air resistance, utilizes the formula s = at²/2, where s is distance, a is gravitational acceleration, and t is time. However, the impact of air resistance complicates the actual height and speed, necessitating more advanced aerodynamic considerations.
PREREQUISITESPhysics students, hobbyist engineers, and anyone interested in the dynamics of projectile motion and aerodynamics, particularly in sports applications.
It would be easy if we could ignore air resistance: It went up for ten seconds, it went down for ten seconds; maximum upwards speed at the moment that it leaves the launcher is equl to the maximum downward speed as it reaches the ground. Now we just have to calculate how high you have to be to take ten seconds to fall to Earth and how fast you're moving when you hit, and ##s=at^2/2## (##s## is the distance, ##a## is the acceleration from gravity, and ##t## is the time) and a bit of algebra will do that.applepi38 said:Years ago, I made a golf ball launcher. It shot straight up at a 90 degree angle. From launch to landing only 2 feet from where it started, it took about 20 seconds before it was back on the ground. Is there a way to determine its maximum speed and height it reached?