Question

in major league baseball, the pitcher's mound is 60 feet from the batter. If a pitcher throws a 92 mph fastball, how much time elapses from when the ball leaves the pitcher's hand until the ball reaches the batter?

Answer 1

0.4464 seconds

Step-by-step explanation:

To determine the time it takes for a baseball to travel from the pitcher's hand to the batter, we need to convert the speed of the fastball from mph (miles per hour) to feet per second.

First, let's convert 92 mph to feet per second:

• 1 mile = 5280 feet
• 1 hour = 3600 seconds

92 mph = (92 * 5280) / 3600 = 134.4 feet per second

Now, we can calculate the time it takes for the ball to reach the batter by dividing the distance between the pitcher's mound and the batter (60 feet) by the speed of the fastball:

Time = Distance / Speed

Time = 60 feet / 134.4 feet per second

Time ≈ 0.4464 seconds

Therefore, approximately 0.4464 seconds elapse from the moment the ball leaves the pitcher's hand until it reaches the batter.

Answer 2

Answer: it takes approximately 0.4464 seconds for the ball to reach the batter.

Step-by-step explanation:

time = distance / speed

Given:

distance = 60 feet

speed = 92 mph

First, we need to convert the speed from mph to feet per second (fps). There are 5280 feet in a mile and 3600 seconds in an hour.

speed in fps = 92 miles/hour * 5280 feet/mile / 3600 seconds/hour

speed in fps = 92 * 5280 / 3600

speed in fps = 134.4 fps

Now, plug this into our formula:

time = 60 feet / 134.4 fps

time = 0.4464 seconds

So, it takes approximately 0.4464 seconds for the ball to reach the batter.