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.
Answer:
Angle = 19.4°
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°