A baseball is seen to pass upward by a window 25 m above the street with a verti

Question

A baseball is seen to pass upward by a window 25 m above the street with a vertical speed of 15 m/s .

Part A

If the ball was thrown from the street, what was its initial speed?

Part B

If the ball was thrown from the street, what altitude does it reach?

Part C

If the ball was thrown from the street, when was it thrown? (That is time elapsed from throwing to reaching the window.)

Part D

If the ball was thrown from the street, when does it reach the street again? (That is time elapsed from the reaching the window on the way up to reaching the street.)

Explanation / Answer

(A) at h = 25 m ,

vf = 15 m/s and a = - 9.8 m/s^2
  
Applying vf^2 - vi^2 = 2 a h

15^2 - v^2 = 2 x -9.8 x 25

v = 26.74 m/s ........Ans

(B) at highest point, v = 0

0^2 - 26.74^2 = 2 x -9.8 x Hmax

Hmax = 36.5 m .......Ans

(C) 15 = 26.74 - 9.8 t

t = 1.20 sec

(D) - 26.74 = 26.74 - 9.8 t

t = 5.46 sec

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