Question

Ryan Howard hit a homerun off the right field foul pole 325 feet from the home plate. The ball hit the bat 3.5 feet above the home plate with an initial velocity of 108 feet/sec and an inclination of 44°. Determine when and the height of the ball when it made contact with the foul pole if there was a 12 feet/sec wind in the same horizontal direction the ball was hit. Show your work.

Answer 1

The height of the ball when it hits the foul pole is approximately 30.6 feet.

Ryan Howard hit a homerun off the right field foul pole 325 feet from the home plate. The ball hit the bat 3.5 feet above the home plate with an initial velocity of 108 feet/sec and an inclination of 44°. Determine when and the height of the ball when it made contact with the foul pole if there was a 12 feet/sec wind in the same horizontal direction the ball was hit. Show your work.

ChatGPT

To solve this problem, we need to use the equations of motion for a projectile motion with air resistance.

First, let's find the initial velocity components:

v₀x = v₀ cos θ = 108 cos 44° ≈ 76.6 ft/s

v₀y = v₀ sin θ = 108 sin 44° ≈ 73.4 ft/s

Next, let's find the time it takes for the ball to reach the foul pole. The horizontal distance to the foul pole is 325 feet, and the initial horizontal velocity is 76.6 ft/s. So the time it takes to reach the foul pole is:

t = d / v₀x = 325 / 76.6 ≈ 4.25 s

During this time, the ball will also be affected by the wind, which has a horizontal velocity of 12 ft/s in the same direction as the ball. So the horizontal distance the ball travels is:

d = (v₀x + vw) t = (76.6 + 12) × 4.25 ≈ 393 ft

Now we can find the vertical distance the ball travels during this time. The initial vertical velocity is 73.4 ft/s, and the acceleration due to gravity is -32.2 ft/s². The vertical position of the ball at any time t is given by:

y = y₀ + v₀y t + 0.5 a t²

where y₀ is the initial height of the ball above the ground. We know that y₀ = 3.5 ft, so we can plug in the values:

y = 3.5 + 73.4 t - 16.1 t²

We want to find the height of the ball when it hits the foul pole, which is at a horizontal distance of 325 feet. So we can use the horizontal distance equation we found earlier to get the time when the ball reaches the foul pole:

325 = (76.6 + 12) t - 0.5 × 0 t²

t ≈ 3.92 s

Now we can plug this value into the vertical position equation to find the height of the ball when it hits the foul pole:

y = 3.5 + 73.4 × 3.92 - 16.1 × (3.92)² ≈ 30.6 ft

So the height of the ball when it hits the foul pole is approximately 30.6 feet.

Answer 2

To solve this problem, we will use the equations of motion for a projectile.

First, we need to find the horizontal and vertical components of the initial velocity:

vx = v * cos θ = 108 * cos 44° = 76.8 ft/s

vy = v * sin θ = 108 * sin 44° = 74.6 ft/s

Next, we can use the vertical motion equation to find the time when the ball reached its maximum height:

y = yo + vyt - 0.5gt^2

0 = 3.5 + 74.6t - 0.532.2t^2 (g = acceleration due to gravity = 32.2 ft/s^2)

16.1t^2 - 74.6t - 3.5 = 0

Using the quadratic formula, we get:

t = 0.27 s or t = 4.27 s

Since the positive root corresponds to the time when the ball was hit, we have:

t = 0.27 s

Next, we can use the horizontal motion equation to find the distance the ball traveled in the horizontal direction:

x = xo + vxt + 0.5at^2

x = 0 + 76.80.27 + 0.500.27^2 (a = acceleration in the horizontal direction = 12 ft/s^2 due to wind)

x = 20.736 ft

Therefore, the ball hit the foul pole 325 - 20.736 = 304.264 ft from the home plate.

Finally, we can use the vertical motion equation to find the height of the ball when it hit the foul pole:

y = yo + vyt - 0.5gt^2

y = 3.5 + 74.60.27 - 0.532.20.27^2

y = 61.2 ft

Therefore, the ball hit the foul pole at a height of 61.2 ft above the home plate.

Explanation: