posted by Abdul .
A ballplayer standing at homeplate hits a baseball that is caught by another player at the same height above the ground from which it was hit. The ball is hit with an initial velocity of 23.0 m/s at an angle of 54.0° above the horizontal.
How high will the ball rise?
. m higher than where it was hit
(b) How much time will elapse from the time the ball leaves the bat until it reaches the fielder?
(c) At what distance from home plate will the fielder be when he catches the ball?
Vo = 23m/s @ 54o.
Xo = 23*cos54 = 13.52 m/s = Hor. component of initial velocity.
Yo = 23*sin54 = 18.61 m/s = Ver. component of initial velocity.
a. Y^2 = Yo^2 + 2g*h.
h = (Y^2-Yo^2)/2g=(0-346.3)/-19.6=17.7 m
b. Y = Yo + g*t.
Tr = (Y-Yo)/g = (0-18.61)/-9.8=1.90 s.
= Rise time.
Tf = Tr = 1.90 s. = Fall time.
T = Tr + Tf = 1.9 +\1.9 = 3.8 s.
c. d = Xo * T = 13.52 * 3.8 = 51.4 m.