Solving for Initial Velocity in Physics Lab | Explanation and Step-by-Step Guide

[daodude1987]

Hey guys, thanks for reading my post. I'm really confused about a lab that was assigned by my teacher. I will try to explain the whole lab the best I can.

Lab: The Balls Lab

Purpose of the Lab: To determine the initial velocity (Vo) and maximum height (dy) of a thrown (and hit) ball

Procedure/Materials:
Materials: tennis ball, tennis racket, stop watch
1. Throw a ball straight up into the air
2. Measure the time the ball is in the air and comes down
3. Due three trials

Data: (here are the 3 time results I got for the ball)
Throwing
Trial 1- 3.21 seconds
Trial 2- 3.10 seconds
Trial 3- 2.94 seconds
Average- 3.08 seconds

Hitting the ball with tennis racket
Trial 1- 1.41 seconds
Trial 2- 2.88 seconds
Trial 3- 1.87 seconds
Average- 2.05 seconds

Conclusion:
Ok here is the problem... I have no clue how to start solving for the initial velocity (Vo) for each of the average result of the ball thrown and hit in the air. At first I thought the initial velocity of the thrown ball was 0 because that is usually the velocity of any object at rest and then going into the air. But my teacher said it wasn't because he wanted the intial velocity from when it left my hand not when it is at rest, which is 0. He also told me that I needed to use one of the equations of motion for constant acceleration such as: Vf=Vo+at, d=1\2(Vf+Vo)t, d=Vo=1/2at(squared), or 2ad=(Vf)squared-(Vo)squared and the acceleration of gravity, -10 m/s to solve for Vo. But he wasn't clear on which one. Also it seems to me that each of these equation require the value of Vo itself! I am also very confused at how to find the height of the thrown ball (dy) because y is not given as a variable in any of these equations.

If you guys can show me how I can solve for the initial velocity (Vo) and the maximum height(dy) of the of the ball thrown using any of the equations above or even your own methods, please enlighten me with you explanation. I am most grateful.

Thank you for taking the time to answer.

 

[jamesrc]

Your procedure should have included stopping the stopwatch when the ball comes down to the exact (or as close as you can get it) position that the ball was in before it was thrown. If you did that, then everything else is no problem. The time to go up is the same as the time to go down, so if you divide the total flight time by 2, you'll have the time it takes to get to the maximum height. You also know that when it gets to the top, its velocity has been decelerated to 0. So use v = vo + a*t here to solve for vo (plug in v = 0, and a = -9.81m/s/s, and t from your experiment divided by 2). To get the max height from there, just use y = .5*a*t^2 + vo*t. You just calculated vo, t is again the time from your experiment divided by 2, and a is -9.81m/s/s again.

 

[M98Ranger]

Hi there! It's great that you are seeking help with your physics lab. Solving for initial velocity can be a bit tricky, but I'll try my best to explain it step-by-step.

Step 1: Understand the concept
Before we dive into solving the equations, it's important to understand the concept of initial velocity. Initial velocity (Vo) is the velocity of an object at the beginning of its motion, before any external forces act on it. In this lab, the initial velocity of the ball will be the velocity with which you throw it into the air.

Step 2: Choose the correct equation
Your teacher mentioned using one of the equations of motion for constant acceleration. In this case, we will use the equation Vf=Vo+at, where Vf is the final velocity, Vo is the initial velocity, a is the acceleration, and t is the time.

Step 3: Plug in the values
Now, let's plug in the values from your data into the equation. Since we are solving for Vo, we will rearrange the equation to be Vo=Vf-at. The final velocity (Vf) will be 0, since the ball will come to a stop at its maximum height. The acceleration (a) will be -10 m/s, as gravity is acting on the ball and causing it to slow down. And the time (t) will be the average time you recorded for the ball to reach its maximum height, which is 3.08 seconds.

Step 4: Calculate Vo
Now, let's plug in the values and solve for Vo. It should look something like this: Vo=0-(-10)(3.08). This will give you an initial velocity of 30.8 m/s.

Step 5: Finding the maximum height
To find the maximum height (dy) of the ball, we can use the equation d=1/2(Vf+Vo)t. Again, Vf will be 0 and Vo will be the value we just calculated, 30.8 m/s. The time (t) will remain the same at 3.08 seconds.

Step 6: Calculate dy
Plug in the values and solve for dy. It should look like this: dy=1/2(0+30.8)(3.08)=47.56 meters.

Step 7: Double check your calculations
To make sure your calculations are correct, you can