Solve Tennis Ball Drop: 0.5kg at 5.6m Height

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A 0.5 kg tennis ball is dropped from a height of 5.6 m, and the discussion focuses on calculating its speed upon contact with the floor. The correct approach involves using the kinematic equation v^2 = u^2 + 2as, where the initial velocity (u) is 0, acceleration (a) is 9.81 m/s², and distance (s) is 5.6 m. There was initial confusion about the placement of variables, particularly misidentifying final velocity as zero. After clarifying the setup, the calculation leads to v^2 = 109.872, and taking the square root provides the final speed. This collaborative problem-solving highlights the importance of correctly identifying variables in physics equations.
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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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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
 
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!)
 
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.
 
Check your equation. If you fill in the numbers you get:

v^2=109.872

You have v squared. Take the square root to get the answer.
 
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!
 
No problem. Be sure to tell your friends to stop by if they need any help (or just want to chat about physics). :smile:
 
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