Question

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

Answer 1

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 + $v_{y}$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

           $v_{y}$ = 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 = $v_{y}$ / 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