A major leaguer hits a baseball so that it leaves the bat at a speed of 28.0 m/s

Question

A major leaguer hits a baseball so that it leaves the bat at a speed of 28.0 m/s and at an angle of 35.3 degrees above the horizontal. You can ignore air resistance. A) At what two times is the baseball at a height of 9.00 above the point at which it left the bat? t1,2 = _____ s B) Calculate the horizontal component of the baseball's velocity at each of the two times you found in part A. v h1,2 = ______ m/s C) Calculate the vertical component of the baseball's velocity at each of the two times you found in part A. v v1,2 = _______ m/s E) What is the direction of the baseball's velocity when it returns to the level at which it left the bat? = _______ degrees below the horizontal

Explanation / Answer

A)28sin35.3 =Vyo

or 9=28sin35.2 t-1/2 gt^2

or t1=0.7712 s and t2=2.58271 s

B)Vx 1,2 =28cos35.3

=22.85 m/s

C)V1=9.21 m/s

V2=-9.21 /s

E)54.7 degrees with the horizontal

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