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.

a small object of mass 3.66 g and charge -17.6 µc is suspended motionless above the ground when immersed in a uniform electric field perpendicular to the ground. what are the magnitude and direction of the electric field?

the answer is a. at the equator