Copyright © Michael Richmond.
This work is licensed under a Creative Commons License.
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.