Amajor-league pitcher can throw a baseball in excess of 41.0 m/s. if a ball is thrown horizontally at this speed, how much will it drop by the time it reaches a catcher who is 17.0 m away from the point of release?

Ball will drop by 0.82 meter.

Explanation:

Horizontal speed = 41 m/s

Horizontal displacement = 17 m

Horizontal acceleration = 0 m/s²

Substituting in s = ut + 0.5at²

    17 = 41 t + 0.5 x 0 x t²

     t = 0.41 s

Now we need to find how much vertical distance ball travels in 0.41 s.

Initial vertical speed = 0 m/s

Time = 0.41s

Vertical acceleration = 9.81 m/s²

Substituting in s = ut + 0.5at²

    s = 0 x 0.41 + 0.5 x 9.81 x 0.41²

    s = 0.82 m

So ball will drop by 0.82 meter.

the order for strong to weak forces such as

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force

the fundamental forces are those forces that act between to objects or element to make interactions. there are four types of fundamental forces such as strong and weak nuclear forces (due to protons and electrons), electromagnetic force (due to the motion of electrons) and gravitational force (due to orbital movement of the planet). if these forces are arranged on the basis of their maximum strength then the order is

strong nuclear force → electromegnetic force → weak nuclear force → gravitational force