- #1
Suyash Singh
- 168
- 1
Homework Statement
https://www.physicsforums.com/attachments/upload_2018-4-29_12-40-30-png.224871/
Homework Equations
The Attempt at a Solution
1/2mv^2=10
1/2m(ucos60)^2=10
1/2mu^2/4=10
mu^2=80
what to do now?
So what was the initial KE? What will it be at height y?Suyash Singh said:Homework Statement
https://www.physicsforums.com/attachments/224871
Homework Equations
The Attempt at a Solution
1/2mv^2=10
1/2m(ucos60)^2=10
1/2mu^2/4=10
mu^2=80
what to do now?
So what was the initial KE?Suyash Singh said:mu^2=80
Yes, so how much KE will be lost in rising height y? And what will the KE be then?Suyash Singh said:Ke + pe = total energy
Your lack of detail has confused you. What exactly is that KE in the first two lines? The KE when?Suyash Singh said:KE-mgy=mgy
KE=2mgy
1/2mv^2=2mgy
the loss in ke is the gain in peharuspex said:Your lack of detail has confused you. What exactly is that KE in the first two lines? The KE when?
That is not what I asked.Suyash Singh said:the loss in ke is the gain in pe
it is the KE at height yharuspex said:That is not what I asked.
You obtained an equation KE=2mgy, where y is the height you are trying to find. But what KE is that? The KE at what point in the process? Check back through your working to make sure.
No.Suyash Singh said:it is the KE at height y
Suyash Singh said:KE(at y)=KE-mgy
KE lost is mgy
Rewrite those statements in a way that makes it clear what KE each of those KE terms refers to. E.g. KE(x) for the KE at height x.Suyash Singh said:KE-mgy=mgy
KE=2mgy
oh sorry i made silly mistakeharuspex said:No.
In posts 8 and 9 you wrote:
Rewrite those statements in a way that makes it clear what KE each of those KE terms refers to. E.g. KE(x) for the KE at height x.
Because this is total KE, made up of horizontal and vertical contributions.Suyash Singh said:how come kinetic energy decreases if horizontal velocity is same
Yes, but you already haveSuyash Singh said:KE at height y
KE(y)=mgy
PE(y)=mgy
(So the initial KE is?)Suyash Singh said:mu^2=80
Suyash Singh said:KE=2mgy
1/2mu^2=40haruspex said:Yes, but you already have
(So the initial KE is?)
And
It is almost always a bad idea to plug in numbers before the final step. Keep everything symbolic as long as you can. I wish now that I had got you to make that switch first.Suyash Singh said:1/2mu^2=40
2mgy=40
y=20/mg
y=2/m
1/4haruspex said:It is almost always a bad idea to plug in numbers before the final step. Keep everything symbolic as long as you can. I wish now that I had got you to make that switch first.
The useful part of what you have so far is that y=E/4g, where E is the initial KE.
Your original calculation of that KE was wrong anyway. You had ½m(u cos 60)2=Efinal. What is cos260?
Quite so, which is a good reason for not plugging that value into equations. You can keep it as a variable name, but it will cancel out later.Suyash Singh said:But he said that 10 joules is useless
haruspex said:Quite so, which is a good reason for not plugging that value into equations. You can keep it as a variable name, but it will cancel out later.
Let's start over...
If the launch speed is u and the mass m, the initial KE, KEi=½mu2.
As you say, the KE at the top, KEf, is 1/4 of that, so mu2/8.
So, what is the PE at the top, PEf?
So what is the height at the top in terms of u and g?
At the point of interest, PEy=KEy=½KEi=¼mu2.
So what is the height of that in terms of u and g?
It was not easy, but perhaps you have learned the benefits of:Suyash Singh said:Omg this question was so hard!
Projectile motion refers to the motion of an object that is launched into the air and then moves under the influence of gravity alone. This type of motion follows a curved path known as a parabola.
Potential energy is the energy an object possesses due to its position or configuration. In projectile motion, potential energy is highest at the highest point of the object's trajectory and decreases as the object moves closer to the ground.
Kinetic energy is the energy an object possesses due to its motion. In projectile motion, kinetic energy is highest at the lowest point of the object's trajectory and decreases as the object moves higher in the air.
Air resistance can affect the trajectory of a projectile by slowing it down and altering its path. This is because air resistance acts in the opposite direction of motion, causing a decrease in the object's velocity and a change in its direction.
Yes, the maximum height and distance of a projectile can be calculated using equations that take into account the initial velocity, angle of launch, and acceleration due to gravity. These equations can vary depending on the specific scenario and the presence of air resistance.