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 24.0 m/s at an angle of 50.0° above the horizontal.
(a) 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?
? m
(a) 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?
? m
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Vertical velocity = 24 . sin 50 = 18.4 m/s
Height reached = v^2 / 2 . g = 18.4^2 / 2 . 9.8 = 17.2 m
b) Time to top height = v / g = 18.4 / 9.8 = 1.88 s
Total time of flight = 2 . t = 3.76 s
c) Distance = v ( horizontal ) . t = 24 . cos 50 . 3.76 = 57.93 = 57.9 m
Height reached = v^2 / 2 . g = 18.4^2 / 2 . 9.8 = 17.2 m
b) Time to top height = v / g = 18.4 / 9.8 = 1.88 s
Total time of flight = 2 . t = 3.76 s
c) Distance = v ( horizontal ) . t = 24 . cos 50 . 3.76 = 57.93 = 57.9 m