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A baseball, hit 3 feet above the ground, leaves the bat at an angle of 45° and is caught by an outfielder 3 feet above the ground and 300 feet from home plate. What is the initial speed of the ball, and how high does it rise? (Round your answers to two decimal places.)

Respuesta :

Answer:

     Initial velocity of ball = 29.92 m/s

     Maximum height reached = 22.82 m

Explanation:

Let the initial velocity be b

Considering vertical motion of base ball:-

Initial velocity, u =  bsin45 = 0.707b

Acceleration , a = -9.81 m/s²

Displacement, s = 0 m

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    0 = 0.707b x t + 0.5 x -9.81 x t²

    t = 0.144b

The time required for the ball to reach out fielder t = 0.144b

Now considering horizontal motion of base ball:-

Initial velocity, u =  bcos45 = 0.707b

Acceleration , a = 0 m/s²

Displacement, s = 300 feet = 91.44 m

Time, t = 0.144b

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    91.14 = 0.707b x 0.144b + 0.5 x 0 x (0.144b)²

    b = 29.92 m/s

Initial velocity of ball = 29.92 m/s

For finding maximum height reached we need to consider  vertical motion of base ball:-

Initial velocity, u =  29.92 x sin45 = 21.16 m/s

Acceleration , a = -9.81 m/s²

Time, t = Half of time of flight = 0.144 x 29.92 x 0.5 = 2.15 s

We have equation of motion s= ut + 0.5 at²

Substituting

   s= ut + 0.5 at²

    s = 21.16 x 2.15 + 0.5 x -9.81 x 2.15²

    s = 22.82 m

Maximum height reached = 22.82 m

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