You throw a baseball directly upward at time =0 at an initial speed of 14.9 m/s. what is the maximum height the ball reaches above where it leaves your hand? ignore air resistance and take =9.80 m/s2.

Maximum height, h = 11.32 meters

Explanation:

It is given that,

The baseball is thrown directly upward at time, t = 0

Initial speed of the baseball, u = 14.9 m/s

Ignoring the resistance in this case and using a = g = 9.8 m/s²

We have to find the maximum height the ball reaches above where it leaves your hand. Let the maximum height is h. Using third equation of motion as :

At maximum height, v = 0

and a = -g = -9.8 m/s²

h = 11.32 meters

Hence, the maximum height of the baseball is 11.32 meters.

it has to be b, about 97 percent.