Kinematics Problem: How Far Will a Pitch Fall Vertically at 61.2 ft?

In summary, the conversation discusses a physics homework problem involving a pitch thrown by Nolan Ryan in 1974. The question asks for the vertical distance the ball would fall by the time it reaches a horizontal distance of 61.2 ft, given its velocity of 100.8 mi/hr. The solution involves converting the velocity to ft/s and using the relevant equation to find the answer. However, the answer range given in the question does not match the calculated answer, leading to a discussion about the use of gravity in ft/s^2 rather than m/s^2.
  • #1
Xaran
2
0

Homework Statement


Directly from my homework.

One of the fastest recorded pitches in major-league baseball, thrown by Nolan Ryan in 1974, was clocked at 100.8 mi/hr. If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically by the time it reached 61.2 ft away?

It wants the answer in ft.

Homework Equations


LaTeX Code: x = x_0 + v_0 t + (1/2) a t^2


The Attempt at a Solution



I started by converting mi/hr to ft/s and got 147.84 ft/s. So with this I figured that the ball would travel the 61.2 ft in 0.41396 seconds. From there I plugged it into the relevant equation to get y = 0 + (0)(0.41396) + (1/2) (-9.8) (0.41396)^2 to get an answer of -0.8934 ft.

The problem is that my question comes with an answer range that says my answer should fall somewhere between 1.8 and 3.6 ft. I've checked and double-checked the problem and I thought I've done it right, but I wanted to get a second opinion. Did I do the problem correctly and the answer range is just wrong, or am I missing something?
 
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  • #2
Xaran said:

Homework Statement


Directly from my homework.

One of the fastest recorded pitches in major-league baseball, thrown by Nolan Ryan in 1974, was clocked at 100.8 mi/hr. If a pitch were thrown horizontally with this velocity, how far would the ball fall vertically by the time it reached 61.2 ft away?

It wants the answer in ft.

Homework Equations


LaTeX Code: x = x_0 + v_0 t + (1/2) a t^2

The Attempt at a Solution



I started by converting mi/hr to ft/s and got 147.84 ft/s. So with this I figured that the ball would travel the 61.2 ft in 0.41396 seconds. From there I plugged it into the relevant equation to get y = 0 + (0)(0.41396) + (1/2) (-9.8) (0.41396)^2 to get an answer of -0.8934 ft.

The problem is that my question comes with an answer range that says my answer should fall somewhere between 1.8 and 3.6 ft. I've checked and double-checked the problem and I thought I've done it right, but I wanted to get a second opinion. Did I do the problem correctly and the answer range is just wrong, or am I missing something?

Welcome to PF.

Shouldn't you be using a different value for gravity in ft/s2
 
  • #3
And there's the small detail I overlooked. Its always something simple. Thanks! And thanks for the welcome. Just started my physics class this semester, so sometimes the little things like that escape me.
 

1. What is the formula for calculating distance in a kinematics problem?

The formula for calculating distance in a kinematics problem is d = v0t + ½at2, where d is the distance, v0 is the initial velocity, t is the time, and a is the acceleration.

2. How do you determine the initial velocity in a kinematics problem?

The initial velocity can be determined by dividing the total distance by the total time and then multiplying it by 2. This is because the average velocity in a constant acceleration scenario is equal to the initial velocity plus the final velocity, divided by 2.

3. What is the acceleration due to gravity?

The acceleration due to gravity is a constant value of 9.8 m/s2 or 32.2 ft/s2, and it is denoted by the letter g. This value is used in kinematics problems involving objects falling due to gravity.

4. How do you calculate the time in a kinematics problem?

The time can be calculated by using the formula t = √(2d/a), where t is the time, d is the distance, and a is the acceleration. This formula is derived from the formula for calculating distance in a kinematics problem.

5. Can you use the same formula for calculating distance in a kinematics problem if the acceleration is not constant?

No, the formula for calculating distance in a kinematics problem can only be used if the acceleration is constant. If the acceleration is not constant, then the area under the velocity-time graph must be calculated to determine the total distance traveled.

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