Kinematics and quadratic formula question

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The discussion revolves around solving a kinematics problem involving a penny thrown straight up at 10 m/s from a height of 1 m. The user calculated the position after 0.5 seconds as 4.775 m and the maximum height as 6.10 m but struggled with finding the speed upon hitting the ground and the total time in the air. They initially obtained a final speed of 9.99 m/s, while the correct answer is 10.9 m/s, prompting a request for clarification. The conversation emphasizes using energy conservation principles and the quadratic formula to accurately determine time and speed, as well as the importance of clearly explaining calculations. The discussion concludes with guidance on calculating average velocity and change in velocity to find the time in the air.
tehreem1
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Homework Statement


You throw a penny straight up at a speed of 10m/s from a height of 1 m.
1. what's it position 0.5 seconds after it is released? - i found that value to be 4.775 m
2. how high up does it go? = 6.10 m
3. what is its speed when it hits the ground? =?
4. how long is the penny in the air? =?

Homework Equations



vf= vi + at
x= .5(vi + vf)t
x= vit + .5 at2
vf2= vi2 + 2ax

The Attempt at a Solution



for 3. i used vf2= vi2 +2ax and i got 9.99 m/s. but i know the correct answer is 10.9 m/s and i am not sure why.
for 4. i am trying to use x= vi * t + .5 at2, but am i supposed to use the quadratic formula to solve for t first?

thankyou!
 
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tehreem1 said:
for 3. i used vf2= vi2 +2ax and i got 9.99 m/s. but i know the correct answer is 10.9 m/s and i am not sure why.
for 4. i am trying to use x= vi * t + .5 at2, but am i supposed to use the quadratic formula to solve for t first?
Welcome to PF.

The first thing you have to do is explain your work. All you have done is given us your answer. We can't read your mind to know why you got 9.99 m/s for final velocity.

It is a good idea to examine the problem first. You know that the initial velocity if 10 m/s up. So when it comes back down to where it started, does it have any more energy than when it left? So what is it speed down when it passes 1 m? Will its speed be more or less one metre further down?

To find the time, you can find the solution to the correct quadratic equation. The correct quadratic equation shows height as a function of time and has the initial height in it somewhere.

AM
 
You can use potential and kinetic energy for some of this. Kinetic energy is the energy it has because it's moving. Potential energy is the energy it has because it's perched up in the air. You can see that an apple that's higher in the tree can make a bigger bruise on your head without anybody throwing it, so we need some kind of energy to represent that, and it's called potential energy.

A falling apple converts its potential energy into kinetic energy, and the formulae for those two energies are carefully designed so that their sum doesn't change. Same applies when you throw it up in the air: it slows down as it gets higher because the kinetic energy gets converted into potential energy.

PE is mass*gravity*height, where gravity means 9.81 m/s^2. KE is mv^2 / 2. Conveniently, mass cancels out of the equation and you can write your master equation:

v^2 / 2 + gh = constant

You get your constant from the initial condition:

10m/s ^2 /2 + 1m * 9.81m/s^2 = C = 59.81 m^2/s^2 = 59.81 Joules/kg
(do check the units please - look up what Joules are)

For instance: maximum height question: it's going nowhere at that moment so KE is zero. The initial KE and PE has turned into all PE. So C = 9.81*height.

Speed when it hits the ground: height is now -1m. Just rearrange the master equation. Notice that v gets squared in the formula for KE so it doesn't matter whether it's going up or down. Either way, the KE is positive.

That method won't help with the other stuff though.

Position after 0.5s: work out the velocity at that time by subtracting 0.5*g from the initial velocity. Average that with the initial velocity to get the average velocity during the 0.5s. Multiply that by 0.5 to get the distance travelled.

Time in the air: you already know the speed when it hits the ground (remember it's negative.) Subtract the initial and final velocity to get the change in velocity. How long does gravity need to make that change?

Adrian.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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