Calculating Time and Distance of a Baseball's Horizontal Motion at 161 km/h

[B-80]

A baseball leaves a pitcher's hand horizontally at a speed of 161 km/h. The distance to the batter is 18.3 m. Neglect air resistance.

(a) How long does it take for the ball to travel the first half of that distance?

161*.27777=44.7m/s

t=m/v=(18.3/2)/44.7=.2 seconds

wrong...

(b) How long does it take for the ball to travel the second half of that distance?
should be the same as (a) right?
(c) How far does the ball fall under gravity during the first half?
can't do without time
(d) How far does the ball fall under gravity during the second half?
can't do without time

 

[Astronuc]

0.2 s looks correct since the ball travels at constant velocity (neglecting air resistance).

Solve b with t = 0.2 s

and then solve c with the appropriate time 0.2 s later.