Solve Tennis Ball Drop: 0.5kg at 5.6m Height

In summary, a tennis ball is dropped from rest at a height of 5.6 m onto a hard floor. The ball has a speed of 9.81 m/s at the instant of contact with the floor. Using the equation v^2=u^2+2*a*s, it is found that the final speed is 109.872 m/s.
  • #1
PhysicslyDSBL
15
0

Homework Statement



A 0.5 kg tennis ball is dropped from rest at a height of 5.6 m onto a hard floor.

Homework Equations



a) What is the speed of the ball at the instant of contact with the floor?

v = m/s

The Attempt at a Solution



I need acceleration and I have attempted using the equation:

v^2 = u^2 +2as to solve for a. I set the variables as follows:

v = 0
u = 9.81
s = 5.6

I got a = 8.6 as my answer.

I either have the wrong equation or I am making a mistake on how I set it up.

I know this is a one-dimension kinematic problem but I have been having trouble setting these up, any help in the right direction is much appreciated!
 
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  • #2
It seems you have some confusion. You used v, the final velocity, as 0. This doesn't make any sense. You used u, the initial velocity, as 9.81, which also doesn't make any sense.

The usual idea for these types of equations is to use a=g=9.81m/s^2, the acceleration due to gravity. Another keyword is "dropped", which means initial velocity is zero. I don't know why you set the final velocity to be 0, but if you read the equation that's what you're trying to solve for!

Let's try this again.

v^2 = u^2 + 2*a*s

u = 0 since the ball is being dropped
a = 9.81m/s^2 since we are on earth
s = 5.6m
v = what we want to solve for
 
  • #3
I think you have your variables screwed up. You know the acceleration a:

a=9.81m/s/s

This is the acceleration due to gravity. It is the same for all free falling bodies close to the Earth's surface.

The final speed v, is not 0, I do not know why you set it as such. This is what your looking for.

The initial speed, u is not 9.81 either. It is zero since the ball starts at rest.

It seems you have your variables screwed up. Does this help clear things up?

(Perfect timing, Nick!)
 
  • #4
Ok, I understand better what I did with the variables, I had the right numbers but not in the right places!

After calculating using:

a = 9.81
u = 0
s = 5.6
v= ?

I get 2(9.81)(5.6) = 109.872 but this is not the final answer, I am unsure where to take this from here but I think it has something to do with the m/s^2.
 
  • #5
Check your equation. If you fill in the numbers you get:

[tex]v^2=109.872[/tex]

You have v squared. Take the square root to get the answer.
 
  • #6
Thank you very much! These are the little mistakes I continually make and it is VERY helpful having a forum like this to help me get it all straight :redface:

You rock!
 
  • #7
No problem. Be sure to tell your friends to stop by if they need any help (or just want to chat about physics). :smile:
 
  • Like
Likes Peter-John King

What is the "Tennis Ball Drop" experiment?

The "Tennis Ball Drop" experiment is a common physics demonstration used to illustrate concepts such as gravity, acceleration, and potential and kinetic energy. It involves dropping a tennis ball of known mass and height and measuring its velocity upon impact with the ground.

How do you set up the "Tennis Ball Drop" experiment?

The experiment requires a tennis ball with a mass of 0.5kg and a height of 5.6m. It can be set up by securing the ball at the top of the 5.6m height and releasing it to fall freely to the ground. A measuring tape or ruler can be used to measure the distance from the ground to the release point.

What is the purpose of the "Tennis Ball Drop" experiment?

The purpose of the experiment is to demonstrate the principles of energy and motion, specifically the relationship between mass, height, and velocity. It can also be used to calculate the acceleration due to gravity on Earth, which is approximately 9.8 m/s².

What factors can affect the results of the "Tennis Ball Drop" experiment?

The results of the experiment can be affected by factors such as air resistance, wind, and the surface the ball lands on. Air resistance can slow down the ball's descent, while wind can alter its trajectory. The surface the ball lands on can also affect its bounce and therefore its final velocity.

How can the "Tennis Ball Drop" experiment be used in real-world applications?

The principles demonstrated in the experiment can be applied to real-world scenarios, such as calculating the trajectory of a projectile or determining the force of impact in a car crash. It also has practical applications in sports, such as understanding the physics behind a tennis ball's trajectory during a serve.

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