Question 48 (III) A baseball is seen to pass upward by a window 28 m above the street with a vertical speed of 13 m/s. If the ball was thrown from the street. (a) what was its initial speed, (b) what altitude does it reach, (c) when was it thrown, and (d) when does it reach the street again?

(a) what was its initial speed?:

(a) what was its initial speed? 28m 13m/s Sidewalk v = 13 m/s a = -9.8 m/ x = 28 m = ? = V + 2ax 1 = V + 2(-9.8)(28) 169 = V + -19.6(28) 169 = V -548.8 +548.8 +548.8 717.8 = V = = 26.79 m/s = 26.79 m/s

(b) what altitude does it reach?:

(b) what altitude does it reach? 28m 13m/s Sidewalk Stopping point of baseball V = 0 m/s a = - 9.8 m / = 26.79 m/s X = ? = V + 2ax = 26.7 + 2(-9.8)x 0 = 716.097-19.6x -716.097 -716.097 -716.097 = -19.6x -19.6 -19.6 X = 36.54 m x = 36.54m

(c) when was it thrown?:

(c) when was it thrown? 28m 13m/s Sidewalk X = 28 m = 26.79 m/s a = -9.8 m/ V = 13 m/s t = ? v = + at 13 = 26.79 + -9.8t -26.79 -26.79 -13.79 = -9.8t -9.8 -9.8 t = 1.407 seconds t = 1.407s

(d) when does it reach the street again? :

(d) when does it reach the street again? 28m 13m/s Sidewalk Stopping point of baseball x = 36.54 m a = 9.8 m/ = 0 m/s t = ? x = t + 1/2a 36.54 = ½(9.8) 36.54 = 4.9 4.9 4.9 7.46 = = t = 2.73 seconds x = 36.54 m v = 0 m/s = 26.79 m/s t = ? v = ½( + v) v = ½(26.79+0) v = 13.395 m/s x = v (t) 36.54 = 13.395(t) t = 2.73 seconds 2.73 s +2.73 s = 5.46 seconds

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