When a ballplayer throws a ball straight up, by how much does the speed of the ball decrease each second while ascending? in the absence of air resistance, by how much does its peed increase each second while it is descending? how much time is required for its ascent as compared to its descent?

Step-by-step explanation:

We are in the case of vertical shot. If we neglect air resistance, the only force acting over the ball is gravity. As gravity acts down to earth and the ball is going up, speed´s ball  will be decreasing by gravity ( 9.8 m/sec²) That means the speed of the ball will be decreasing 9.8 m/sec, each second.

When the ball get the maximum height  Vy = 0 then the ball stars descending, and when the ball is descending again gravity is the only force acting gravity acts in the same direction as the ball´s movement, so the ball´s speed will increase at a rate of 9.8 m/sec  each second.

It is a well known fact that the time of the ball to get maximum height is the same as for the ball to fall down to earth. In other words time from the beginning of the movement up to maximum height is equal to time for the ball to fall down.

1 inch

Step-by-step explanation:

If we call the thickness of the frame x, then the dimensions of the glass are:

w = 8 − 2x

h = 10 − 2x

The area of the glass is width times height:

48 = (8 − 2x) (10 − 2x)

Distribute and simplify:

48 = 80 − 16x − 20x + 4x²

0 = 4x² − 36x + 32

0 = x² − 9x + 8

Factor and solve:

0 = (x − 1) (x − 8)

x = 1 or x = 8

x can't be more than 4 (half the width of the frame), so x = 1 inch.