The Rogers Centre in Toronto, Ontario, has a center field fence that is feet high and feet from home plate. A ball is hit feet above the ground and leaves the bat at a speed of miles per hour. Determine analytically the minimum angle required for the hit to be a home run.

The question is not complete and the complete question is;

The Rogers Centre in Toronto, Ontario, has a center field fence that is 10 feet high and 400 feet from home plate. A ball is hit 3 feet above the ground and leaves the bat at a speed of 100 miles per hour. Determine analytically the minimum angle required for the hit to be a home run.

Angle = 19.4°

Step-by-step explanation:

In model for projectile motion, we know that;

→r(t) = (VoCos θ)tî + [h + (Vosinθ)t - 1/2gt²] ĵ

In the question, h= 3;

g in ft/s² = 32 ft/s² and Vo = 100 miles per hour

Let's convert Vo to ft/s

1 mile/h = = 1.467 ft/s

So, 100 miles/h = 100 x 1.4667 ft/s = 146.67 ft/s

Plugging it into the →r(t) equation to obtain;

→r(t) = (146.67Cos θ)tî + [3 + (146.67sinθ)t - (32/2)t²] ĵ

Using the general form of;

→r(t) = xî + yĵ

Thus; x = (146.67Cos θ)t and

y = [3 + (146.67sinθ)t - (32/2)t²]

In x; t = x/(146.67 Cosθ)

So plugging this value of t into the equation for y, we obtain;

y = [3 + [(146.67sinθ) (x/(146.67 Cosθ))] - [(16)(x/(146.67 Cosθ)²]

y = 3 + xtanθ - 16x²/(146.67 Cosθ)²

From the question, our y = the height of the field fence while x is the distance from the home plate.

Thus, y = 10 ft and x = 400ft.

Plugging them into the equation, to obtain;

10 = 3 + 400tanθ - 16(400)²/(146.67 Cosθ)²

10 - 3 = 400tanθ - (119/Cos²θ)

7 = 400tanθ - (119/Cos²θ)

From trigonometric identities, we know that, 1/Cos²θ = Sec²θ

Thus,

7 = 400tanθ - 119Sec²θ

Again from trigonometric identities,

Sec²θ = 1 + Tan²θ

Thus 7 = 400tanθ - 119(1 + Tan²θ)

Arranging this to get;

119tan²θ - 400tanθ + 126 = 0

Using quadratic equation to solve,

tanθ = 0.3518

So,θ = tan^(-1) 0.3518

So,θ = 19.382° or approximately, 19.4°

hope this !

step-by-step explanation:

d = distance between moon and sun, which in this problem is the hypotenuse of a right triangle.  

the distance y is the side opposite to angle x in the right triangle.   use the definition of the sine function from basic trigonometry.

sin(x) = y/d ⇒

d = y/sin(x)

guess at question:

a vertical line intersects the graph of g(x,y) = 0 at two points. what conclusion follows about the function f(x) implicitly defined by g(x,f(x)) = 0?

the domain of f(x) cannot include the x coordinate of any such vertical line.

informally, f(x) is not a function, and

g(x,y)=0 cannot be solved for y uniquely.

the relation r = {(x,y) such that g(x,y) = 0} is a function if

for all (a,b) ∊ r and (c,d) ∊ r, a=c implies b=d.

step-by-step explanation: