A baseball is hit into the air at an initial speed of 36.6m/sand an angle of 50.

Question

A baseball is hit into the air at an initial speed of 36.6m/sand an angle of 50.0o above the horizontal. At thesame time, the center fielder starts running away from the batterand catches the ball 0.914m above the level at which it washit. If the center fielder is initially 1.10*102mfrom home plate, what must be his average speed? answer: 4.2m/s A baseball is hit into the air at an initial speed of 36.6m/sand an angle of 50.0o above the horizontal. At thesame time, the center fielder starts running away from the batterand catches the ball 0.914m above the level at which it washit. If the center fielder is initially 1.10*102mfrom home plate, what must be his average speed? answer: 4.2m/s

Explanation / Answer

Write the two kinematics equations. Each is   final position = initial position + init velocity* time + (1/2) a t2 . horizontal:     d = 0 + 36.6 cos50 t   + 0 . vertical:     0.914 = 0 + 36.6 sin50  t - (1/2) 9.8 t2 . Use the vertical equation to find t, the time the ballis in the air. It is a quadratic equation, so... .          4.9t2    -   28.0373 t + 0.914 = 0 .         t =    5.689 seconds . Now find d: .     d = 36.6 * cos50 * 5.689 =    133.84 meters      this is the horiz distance theball traveled. . The fielder has to run     133.84 - 110   =   23.84 meters . So his speed must be    distance / time = 23.84 /5.689 =     4.19  or   4.2   m/s . Use the vertical equation to find t, the time the ballis in the air. It is a quadratic equation, so... .          4.9t2    -   28.0373 t + 0.914 = 0 .         t =    5.689 seconds . Now find d: .     d = 36.6 * cos50 * 5.689 =    133.84 meters      this is the horiz distance theball traveled. . The fielder has to run     133.84 - 110   =   23.84 meters . So his speed must be    distance / time = 23.84 /5.689 =     4.19  or   4.2   m/s

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