Playing shortstop, you pick up a ball and throw it to second base. the ball is thrown horizontally, with a speed of 21m/s, directly towards point a. when the ball reaches the second baseman 0.46 s later, it is caught at point b. part a.) how far were you from the second baseman? part b.) what is the distance of vertical drop, from a to b

distance to second baseman = 9.66 m

vertical drop = 1.058m

Explanation:

A) applying motion equation to horizontal motion

(assuming no acceleration in horizontal motion)

distance to second baseman = velocity * time

                                                = 21 * 0.46

                                                = 9.66 m

B) applying motion equation to vertical motion

(starting velocity was horizontal. therefore no vertical component. subjected to gravity ,g=10m/s2)

vertical drop = 1.058m

that would be because of both gravity and friction. so     c