A person throws a ball up into the air, and the ball falls back toward earth at which point where the kinetic energy be the lowest
The kinetic energy of the ball will be the lowest at the highest point of its trajectory, just before it starts falling back toward Earth. At this point, the ball has reached its maximum height and its velocity has decreased to zero. Since kinetic energy (KE) is given by the equation KE = 1/2mv^2, where m is the mass of the ball and v is its velocity, when the velocity is zero, the kinetic energy is also zero.
Who said there was no horizontal speed?
If it is not thrown exactly straight up, it has horizontal speed u and vertical speed u
Since there is no horizontal force u is constant for the whole trip.
The vertical speed though Vi at the bottom goes to zero at the top and then down to - Vi back at the bottom
the total speed at the top is therefore minimum and the minimal Ke there is (1/2) m u^2
Apologies for the confusion in my previous response. You are correct; I made an oversight by not considering the horizontal speed (u) in the ball's motion.
In this case, when the ball is thrown with an initial horizontal speed u, the vertical motion and horizontal motion are independent of each other. The vertical motion is affected by gravity, while the horizontal motion remains unaffected as there is no horizontal force acting on the ball.
As the ball reaches its maximum height, the vertical speed (Vy) becomes zero, while the horizontal speed (u) remains constant. At this point, the total speed (V) is given by the Pythagorean theorem: V = √(u^2 + Vy^2). Since Vy is zero, the total speed V at the highest point is equal to u, and the kinetic energy will be minimum there, as you correctly stated.
Therefore, the lowest kinetic energy will be at the maximum height of the ball, where the total speed is minimum. The specific value of the kinetic energy will be (1/2) mu^2, where m is the mass of the ball and u is its initial horizontal speed.