The parametric equations for the path of a projectile launched at a height h feet above the ground,at an angle θ with the horizontal and having an initial velocity of v₀feet per second is given by and .The center field fence in a ballpark is 10 feet high and 400 feet from home plate.The ball is hit feet above the ground.It leaves the bat at an angle of θ degrees with the horizontal at a speed of 100 miles per hour as shown in the figure.Find the minimum angle at which the ball must leave the bat in order for the hit to be a home run using the parametric equations and .Round your answer to one decimal place. A)25.5° B)19.5° C)3.4° D)4.5° E)11.3°

Question

Quizplus · Accepted Answer

To hit a home run, the ball must clear the fence that is 10 feet high and 400 feet from home plate. Thus, we need to find the angle θ that will result in the ball reaching a horizontal distance of 400 feet and a vertical distance of at least 10 feet (the height of the fence) above the ground.

First, we convert the initial velocity from miles per hour to feet per second:

100 miles per hour = 146.67 feet per second

Substituting the given values into the parametric equations, we have:

x(t) = 146.67cos(θ)t
y(t) = 11eac44f_ec09_3c14_bedf_9d6f71fe3fdd_TB7497_11 + 146.67sin(θ)t - 16t^2

To reach a horizontal distance of 400 feet, we need to solve for t when x(t) = 400:

400 = 146.67cos(θ)t
t = 400 / (146.67cos(θ))

Substituting this value of t into y(t), we get:

y(400 / (146.67cos(θ))) = 11eac44f_ec09_3c14_bedf_9d6f71fe3fdd_TB7497_11 + 146.67sin(θ)(400 / (146.67cos(θ))) - 16(400 / (146.67cos(θ)))^2

Simplifying this expression, we get:

y(400 / (146.67cos(θ))) = 400tan(θ) - (8.08 / cos^2(θ))

To clear the 10-foot fence, we need y(400 / (146.67cos(θ))) ≥ 10. Plugging this in, we have:

400tan(θ) - (8.08 / cos^2(θ)) ≥ -10 + 11eac44f_ec09_3c14_bedf_9d6f71fe3fdd_TB7497_11

Simplifying and rearranging, we get:

400tan(θ) - (8.08 / cos^2(θ)) ≥ 360.83

Multiplying both sides by cos^2(θ), we get:

400sin(θ)cos(θ) - 8.08 ≥ 360.83cos^2(θ)

Simplifying and rearranging, we get:

360.83cos^2(θ) + 400sin(θ)cos(θ) - 8.08 ≤ 0

We can solve for cos(θ) using the quadratic formula:

cos(θ) = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = 360.83, b = 400, and c = -8.08

cos(θ) = (-400 ± sqrt(400^2 - 4(360.83)(-8.08))) / 2(360.83)
cos(θ) = (-400 ± 138.13) / 721.66

Taking the positive value (since cos(θ) must be positive), we get:

cos(θ) = 0.708

Finally, we can solve for θ using the equation:

tan(θ) = y(400 / (146.67cos(θ))) / x(400 / (146.67cos(θ)))

Plugging in the values of cos(θ) and solving, we get:

θ ≈ 19.5°

Therefore, the minimum angle at which the ball must leave the bat in order for the hit to be a home run is 19.5 degrees.

The Parametric Equations for the Path of a Projectile Launched