A pitcher throws a baseball horizontally from the mound to home plate. The ball falls 0.868 m (2.85 ft) by the time it reaches home plate 18.3 m (60 ft) away. How fast was the pitcher's pitch?

Well, if the pitcher threw the baseball horizontally, it means that gravity didn't have any effect on the ball's horizontal motion. So, let's just focus on the ball's initial velocity.

To find the initial velocity, we need to consider the horizontal distance the ball traveled. Using the equation: distance = velocity × time, we can rearrange it to: velocity = distance / time.

In this case, the distance is 18.3 m (60 ft) and the time is the same for both the horizontal and vertical components. So, the velocity is 18.3 m / time.

Now, let's focus on the vertical distance traveled by the ball. We know that the ball fell 0.868 m (2.85 ft) during this time. Using the equation for free fall distance: distance = 0.5 × acceleration × time^2, we can rearrange it to: time = sqrt((2 × distance) / acceleration).

For the vertical motion, the acceleration is the acceleration due to gravity, which is approximately 9.8 m/s^2.

Plugging in the numbers, we have time = sqrt((2 × 0.868 m) / 9.8 m/s^2).

Once you find the time, you can use it to calculate the horizontal velocity as velocity = 18.3 m / time.

So, my non-vertical answer is: the speed of the pitcher's pitch was approximately 18.3 m divided by the time it took for the ball to fall 0.868 m, which is approximately 18.3 m/s. But hey, don't worry, I'm sure I didn't drop the ball on this one!

To find the speed of the pitcher's pitch, we can use the horizontal distance traveled by the ball and the vertical distance it dropped.

Given:
- Horizontal distance (d): 18.3 m
- Vertical distance (h): 0.868 m

We can use the following formula to calculate the speed (v):

v = √(d * g * (2h + d)) / d

where g is the acceleration due to gravity, approximately 9.8 m/s².

Substituting the values into the formula:

v = √(18.3 * 9.8 * (2 * 0.868 + 18.3)) / 18.3

Calculating this expression step-by-step:

v = √(18.3 * 9.8 * (1.736 + 18.3)) / 18.3

v = √(18.3 * 9.8 * 20.036) / 18.3

v = √(3610.62848) / 18.3

v = √3610.62848 / 18.3

v = 60.067 / 18.3

v ≈ 3.28 m/s

Therefore, the speed of the pitcher's pitch was approximately 3.28 m/s.

To determine the speed of the pitcher's pitch, we need to consider the horizontal distance and the vertical distance the ball traveled.

In this case, we know that the vertical distance is 0.868 m (2.85 ft). Since the ball was thrown horizontally, there is no vertical velocity initially. Therefore, we can assume that the vertical distance traveled is due to the force of gravity.

Using the equation for free-fall distance, we have:

d = (1/2) * g * t^2

where:
d is the vertical distance (0.868 m)
g is the acceleration due to gravity (9.8 m/s^2)
t is the time it takes for the ball to reach the home plate

Solving for t, we get:

t = sqrt((2 * d) / g)

Substituting the given values, we have:

t = sqrt((2 * 0.868) / 9.8) ≈ 0.167 seconds

Next, we can use the time and horizontal distance to find the speed (velocity) of the pitch.

The horizontal distance traveled is 18.3 m (60 ft). Since the ball was thrown horizontally, we can assume that there is no horizontal acceleration acting on it (in ideal conditions).

Using the equation for velocity (v = d / t), we have:

v = 18.3 m / 0.167 s ≈ 109.49 m/s

Therefore, the speed of the pitcher's pitch is approximately 109.49 m/s.

Answer to my own question, you use d=vi*t+1/2*a*t^2 after solving for t you use d=vxo*t