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3 years ago

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

Physics

1 answer:

Alex Ar [27]3 years ago

3 0

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

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