Answer to Question #85204 in Physics for Maria

Question

A baseball pitch leaves the pitcher's hand horizontally at a velocity of 180 km/h .
By what % will the pull of gravity change the magnitude of the velocity when the ball reaches the batter, 18 m away? For this estimate, ignore air resistance and spin on the ball.

Answer 1

Initial velocity:

v=v_x=180 km/h=180/3.6 m/s=50 m/s.

v_y=gt.

x=v_x t→t=x/v_x .

v_y=g x/v_x

New velocity:

V=√(v_x^2+v_y^2 )=√((v_y )^2+(v_x )^2 )

% the pull of gravity changes the magnitude of the velocity when the ball reaches the batter:

((V-v))/v=√((v_y )^2+(v_x )^2 )/v_x -1=√(1+(v_y/v_x )^2 )-1=√(1+(g x/(v_x^2 ))^2 )-1

((V-v))/v=√(1+(9.8 18/〖50〗^2 )^2 )-1=0.0025

It is 0.25%.