The parametric equations for the path of a projectile launched at a height h feet above the ground, at an angle with the horizontal and having an initial velocity of feet per second is given by and . The center field fence in a ballpark is feet high and feet from home plate. The ball is hit 3 feet above the ground. It leaves the bat at an angle of 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.

Question

The parametric equations for the path of a projectile launched at a height h feet above the ground, at an angle  with the horizontal and having an initial velocity of  feet per second is given by  and  . The center field fence in a ballpark is  feet high and  feet from home plate. The ball is hit 3 feet above the ground. It leaves the bat at an angle of  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.

Quizplus · Accepted Answer

First, we need to find the equation of the path of the projectile. We are given that it is launched at a height of 3 feet, at an angle of 20 degrees with the horizontal, and with an initial velocity of 100 mph. We will assume that air resistance is negligible, so the only force acting on the ball is gravity.

Using the formulas for the x and y components of the velocity and position of a projectile, we get:

x(t) = 176.0t
y(t) = -16t^2 + 34.6t + 3

where x and y are in feet and t is in seconds.

Next, we need to find the point at which the ball hits the center field fence. The fence is 400 feet away from home plate and 10 feet high, so its equation is:

y = 10

Setting y(t) = 10 and solving for t, we get:

-16t^2 + 34.6t + 3 = 10
-16t^2 + 34.6t - 7 = 0
t = 0.56 or t = 2.06

We can discard the solution t = 0.56 because it corresponds to the ball being below the ground, which is impossible. Therefore, the ball hits the center field fence at t = 2.06 seconds.

Substituting t = 2.06 into x(t) and y(t), we get:

x(2.06) = 362.56
y(2.06) = 9.94

So the ball just barely misses the center field fence.

Finally, we need to find the minimum angle at which the ball must be hit to clear the fence (i.e., to have y(t) > 10 for some value of t). Since the ball just barely misses the center field fence, we can assume that it will clear the fence if it is hit at a slightly higher angle.

To find this angle, we can use the fact that the maximum range of a projectile (i.e., the distance traveled in the x direction) occurs when it is launched at an angle of 45 degrees. In our case, we want to find the angle at which the ball just clears the fence, so we can solve for the launch angle that gives a range of exactly 400 feet.

The formula for the range of a projectile launched at an angle  is:

R = v0^2 sin(2θ) / g

where R is the range, v0 is the initial velocity, θ is the launch angle, and g is the acceleration due to gravity.

We want R = 400 feet, so we can solve for θ:

400 = (100^2 / 32.2) sin(2θ)
sin(2θ) = 1.244

Since sin(2θ) cannot be greater than 1, there is no solution to this equation. Therefore, the ball cannot be hit at an angle that clears the center field fence.

The Parametric Equations for the Path of a Projectile Launched