Consider a projectile launched at a height h feet above the ground and at an ang

Question

Consider a projectile launched at a height h feet above the ground and at an angle theta with the horizontal. If the initial velocity is y0 feet per second, the path of the projectile is modeled by the parametric equations x = and y = h + (v0 sin(theta))t - l6t^2. The center field fence in a ballpark is 10 feet high and 400 feet from home plate. The ball is hit h = 2 feet above the ground. It leaves the bat at an angle of 8 degrees with the horizontal at a speed of 101 miles per hour (see figure). (a) Write a set of parametric equations for the path of the ball. Write your equations in terms of t and theta.) (a) Write a set of parametric equations for the path of the ball. (Write your equations in terms of t and theta.) (b) Use a graphing utility to graph the path of the ball when theta = 15 degree. Is the hit a home run? Yes No (c) Use a graphing utility to graph the path of the ball when theta = 23 degree. Is the hit a home run? Yes No (d) Find the minimum angle at which the ball must leave the bat in order for the hit to be a home run. (Round your answer to one decimal place.)

Explanation / Answer

For it to be a homerun, we need to find theta such that x > 400 and y > 10

We have x = (2222/15)cos(theta)*t

And that must be greater than or equal to 400

Now, for minimized angle, let x = 400

(2222/15)cos(theta)*t = 400

t = 2.7002700270027003 / cos(theta)

y = 2 + (2222/15)sin(theta)*t - 16t^2

And this must be equal to 10.

2 + (2222/15)sin(theta)*t - 16t^2 > 10

16t^2 - (2222/15)sin(theta)*t + 8 = 0

Now, plug in for t with 2.7002700270027003 / cos(theta) :

16(2.7002700270027003 / cos(theta))^2 - (2222/15)sin(theta)*2.7002700270027003 / cos(theta) + 8 = 0

116.6633315sec^2(theta) - 400tan(theta) + 8 = 0

116.6633315(1 + tan^2(theta)) - 400tan(theta) + 8 = 0

116.6633315 + 116.6633315tan^2(theta) - 400tan(theta) + 8 = 0

116.6633315tan^2(theta) - 400tan(theta) + 124.6633315 = 0

This is a quadratic in tan(theta), which we now solve

tantheta = (400 +/- sqrt(160000 - 4(116.6633315)(124.6633315)) / [2(116.6633315)]

tantheta = (400 +/- 319.1) / 233.326663

tan(theta) = 3.0819452468661929 , tan(theta) = 0.3467241975684536318937540370171925014844960089

Solving for principal solutions, we get :

theta = 72.02 degrees , theta = 19.1 degrees

So, answer :

19.1 degrees ------> ANSWER

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