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A major league baseball pitcher throws a pitch that follows these parametric equations:

x(t) = 147t
y(t) = –16t2 + 4t + 5.
The time units are seconds and the distance units are feet. The distance between the location of the pitcher and homeplate (where the batter stands) is 60.5 feet. Give EXACT answers, unless instructed otherwise.

(d) Recall that the speed of the baseball at time t is

s(t)=√ [x '(t)]2 + [y ' (t)]2ft/sec.

What is the speed of the baseball (in mph) when it passes over homeplate?

___________ mph

(e) At what time does the baseball hit the ground, assuming the batter and catcher miss the ball? ______ sec.

(f) What is the magnitude of the angle at which the baseball hits the ground? ________ rad.

(This is the absolute value of the angle between the tangential line to the path of the ball and the ground. Give your answer in radians to three decimal places.)

Respuesta :

Answer:

d)    v = 100.2 mph, e) t = 1.25 s, f) -0.0340

Explanation:

d) This is an exercise that we can solve using projectile launch equations, let's start by calculating the time the ball will take to take home-plate

          .x = 147 t

          t = x / 147

          t = 60.5 / 147

          t = 0.41156 s

Let's use Pythagoras' theorem to find the speed

           v = √ vₓ² + [tex]v_{y}[/tex]²

           vₓ = dx / dt

           vₓ = 147

           v_{y} = dy / dt

            v_{y} = 4 -16 t

We look for speed for the time of arriving at home

           [tex]v_{y}[/tex] = 4 - 16 0.41156

            v_{y} = -2,585 ft / s

            v_{y} = -2.585 ft/ s ( 1 mile /5280 foot) (3600s/1h)

           

Let's calculate the speed

             v = √ (147² + 2,585²)

             v = 147.02 ft / s

              v =  147.0 ft/s (1 mile/5280 feet)(3600s/1h)

             v = 100.2 mph

e) the time it takes for the ball to reach the floor and = 0 foot

           

       y = 5 + 4 t - 16 t²

       0 = 5 + 4t - 16t²

       t² –t / 4 -5/4 = 0

       t² -0.25 t -1.25 = 0

We solve the equation and second degree

       t = [0.25 ±√(0.25² + 4 1.25)] / 2

       t = [0.25 ± 2.25] / 2

       t₁ = 1.25 s

       t₂ = -1 s

The positive time is correct

       t = 1.25 s

f) The angle of speed when the ball passes home

         tan θ = [tex]v_{y}[/tex] / vₓ

         θ = tan⁺¹ (v_{y} / vx)

         

The distance x is given in the exercise

          x = 60.5 foot

          vₓ = 147 foot / s

           

The speed y is t = 1.25 s

          v_{y} = 5 + 4 1.25 - 16 1.25²

          v_{y} = -15 foot / s

         

         θ = tan⁻¹ (-15/147)

         θ = -1,947º = -0,0340 rad

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