Kinematics and quadratic formula question

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SUMMARY

The discussion focuses on solving kinematics problems involving a penny thrown vertically with an initial speed of 10 m/s from a height of 1 m. The position after 0.5 seconds is calculated to be 4.775 m, and the maximum height reached is 6.10 m. The final speed upon hitting the ground is determined using the equation vf² = vi² + 2ax, with the correct answer being 10.9 m/s. To find the time the penny is in the air, the quadratic formula is suggested to solve for t in the equation x = vit + 0.5at².

PREREQUISITES
  • Understanding of kinematic equations: vf = vi + at, x = vit + 0.5at², vf² = vi² + 2ax
  • Knowledge of potential and kinetic energy concepts
  • Ability to apply the quadratic formula to solve equations
  • Familiarity with gravitational acceleration (9.81 m/s²)
NEXT STEPS
  • Study the derivation and application of kinematic equations in vertical motion
  • Learn how to apply the quadratic formula to solve for time in projectile motion problems
  • Explore the relationship between potential and kinetic energy in motion
  • Practice solving real-world problems involving vertical motion and energy conservation
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Students studying physics, educators teaching kinematics, and anyone interested in understanding the principles of motion and energy conservation in vertical throws.

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.
 

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