Question

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.

Answer 1

Maximum height, h = 11.32 meters

Step-by-step 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 :

`v^2-u^2=2ah`

At maximum height, v = 0

and a = -g = -9.8 m/s²

`h=(v^2-u^2)/(2a)`

`h=(0-(14.9\ m/s)^2)/(2* -9.8\ m/s^2)`

h = 11.32 meters

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