When a ball player throws a ball straight up?

Question

When a ball player throws a ball straight up, by how
much does the ball's speed decrease each second during
the ascent? In the absence of air resistance, by how
much does the speed increase each second the ball is
descending? How much time, from the starting position,
is required for rising compared to falling?

Answer 1

i am just guessing here but i think it is something like 9.81.

Answer 2

In the absence of air resistance, the speed of a ball thrown vertically upwards decreases at a rate of 9.8 m/s each second, i.e., it decelerates at a rate of 9.8 ms^-2, which is due to resistance by the Earth's gravity. Similarly, when coming down, the speed of ball increases at a rate of 9.8 m/s, i.e., it accelerates at a rate of 9.8 ms^-2. The same time is required for both ascending n descending.

And since ball players have only been found on Earth till date, i am calculating as per Earth's gravity.

Answer 3

During ascent, the ball slows down at the rate of 9.8 feet per second per second, or 32.2 feet per second squared, if air resistance is ignored. The time in rising is equal to the time in falling. To figure the time involved, you have to know the velocity at which the ball left the players hand. V=1/2 at². a = 32.2 feet/sec². Plug in V and solve for t² and then take the square root.