Question 1:
Joe throws a ball upwards at 15 m/s. How long until the ball falls back into his hand?
Answer: There are two ways to solve the problem. Method 1: figure out how long it takes for the ball to stop moving upwards (at the peak of its trajectory), then double that time. Know initial velocity vo = 15 m/s up final velocity v = 0 m/s at peak acceleration a = -9.8 m/s^2 Calc v = vo + a * t 0 m/s = 15 m/s + (-9.8 m/s^2) * t -15 m/s t = ----------- = 1.53 sec -9.8 m/s^2 --> total time = 2 * t = 3.06 sec Method 2: Calculate the time it takes for the ball to have a total displacement of zero: Know: displacement (x - xo) = 0 m initial velocity vo = 15 m/s acceleration a = -9.8 m/s^2 (x - xo) = vo * t + 0.5 * a * t^2 0 m = (15 m/s) * t - (4.9 m/s^2) * t^2 0 m/s = (15 m/s) - (4.9 m/s^2) * t -15 m/s ---> t = ------------- = 3.06 sec -4.9 m/s^2
Copyright © Michael Richmond.
This work is licensed under a Creative Commons License.