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.

Question

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? 
s

(c) At what distance from home plate will the fielder be when he catches the ball?

Henry · Answer

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.