If I throw a ball straight up into the air, we say the ball is a "freely falling object" on its way up, on its way back down, or both on its way up and on its way back down.
A ball tossed vertically upward rises, reaches its highest point, and then falls back to its starting point. During this time, the acceleration of the ball is always in the direction of motion, opposite its velocity, directed downward, or directed upward.
I throw a ball across an open field. At what part of its path does the ball have a minimum speed? right before it hits the ground halfway to the top at the top of its path right after it leaves my hand There's not enough information to say.
Ralph asked me a question the other day. Consider a car accelerating forward. Its acceleration is 1.8 m/s2. During the first second, the car accelerates from 0 to 1.8 m/s. Ralph thought that since the velocity at the end of the first second is 1.8 m/s, the car would travel 1.8 m during that first second. But someone told him that the answer is actually 0.9 m. Can you help Ralph understand why? Don't just say, "Because the formula in the book says so." The answer would be 1.8 m if that were the velocity of the car during the entire second. But that was its speed only at the end of the second. Since it started from rest (v=0), the average velocity during the first second is (0+1.8)/2=0.9 m/s. So the distance traveled is 0.9 m. That's why there is a factor 1/2 in the formula (1/2)at^2. We calculate the distance from the average velocity over the time interval.