A baseball is seen to pass upward by a window with a vertical speed of 13 m/s .

Question

A baseball is seen to pass upward by a window with a vertical speed of 13 m/s . The ball was thrown by a person 15 m below on the street. A) What was its initial speed? Express your answer to two significant figures and include the appropriate units. B) What altitude does it reach? Express your answer to two significant figures and include the appropriate units. C) What time elapsed since it was thrown? Express your answer to two significant figures and include the appropriate units. D) What time will it take the baseball to reach ground again, counting from the moment it passed the window upward? Express your answer to two significant figures and include the appropriate units.

Explanation / Answer

given data:

h = 15m

v = 13m/s

(a)

v2 = u2 + 2ax1

169 = u2 + 2(-9.8)(15)

169 = u2 – 294

u2 = 463

u =21.517 m/s

u = 21 m/s (in two significant figures)

(b)

v2 = u2 + 2ax1

0 = (21.517)2 + 2(-9.8)x

0 = 463 - 19.6x

-716.097 = -19.6x

x = 23.622m

x = 23 m

(c)

v = u + at

13 = 21.517 + -9.8t

13.8 = 9.8t

t = 0.87 second

(d)

x = ut + ½ at2

-23.622 = ½*(9.8)*t2

t = 2.1956 seconds

total time = 2.1956 s +2.1956 s

t = 4.4 seconds

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