Projectile motion helicopter

[BiGyElLoWhAt]

but if i will take initial velocity as zero all -maximum height,range,time of flight would come out as zero because all of them have initial velocity in their formula.

No it doesn't. Write the equations of motion please.

 

[cjl]

Not if you start at some nonzero initial height h₀.

 

[gracy]

No it doesn't. Write the equations of motion please.

equation for Range =u square sin2theta/g
time of flight=2u sin theta/g
maximum height=u square sin square theta/2g
here u=initial velocity
what i am missing?
is u not equal to initial velocity.then u is what ?

 

[cjl]

Are those the only form of the equations you were given (or have in your book)? Those implicitly assume a starting height of zero, which is incorrect both for the quick example that BiGyElLoWhAt and I are trying to get you to work, and for the problem stated in the original post.

 

[gracy]

Those implicitly assume a starting height of zero, which is incorrect

is it incorrect or correct but not the only case of the projectile?

 

[cjl]

Your equations are correct for a special case (namely, where the starting height is zero), which does not apply to the problem as stated in the original post (or the problem we want you to work out).

 

[gracy]

Are those the only form of the equations you were given (or have in your book)? Those implicitly assume a starting height of zero, which is incorrect both for the quick example that BiGyElLoWhAt and I are trying to get you to work, and for the problem stated in the original post.

oh i got your point .give me some time i will work it out by myself.

 

[BiGyElLoWhAt]

I have no idea what you're writing. The position for an object at any point in time (with constant acceleration) is given by:
$x= x_0 + v_{(0\ of\ x)} t + \frac{1}{2}a_xt^2$
y is the same, but with y's.

 

[gracy]

I have no idea what you're writing. The position for an object at any point in time (with constant acceleration) is given by:
$x= x_0 + v_{(0\ of\ x)} t + \frac{1}{2}a_xt^2$
y is the same, but with y's.

i am sorry but i didn't understand your post.

 

[cjl]

i am sorry but i didn't understand your post.

Do you have any other equations than the ones you mentioned above? As I said, the ones you already posted won't work for what we want to do right now.

 

[gracy]

Do you have any other equations than the ones you mentioned above? As I said, the ones you already posted won't work for what we want to do right now.

no,i don't have.i have given and taught these only.i want to learn and explore more.. i tried.as far as i know
the range is defined as the distance between the launch point and the point where the projectile hits the ground.so in this vertical free fall projectile it (range) should be simply height from where the ball is dropped.right?

 

[cjl]

Range is defined as the horizontal distance from launch point to impact point, so for a vertical free fall, range is zero. Was the problem in the original post given to you in class?

 

[gracy]

Was the problem in the original post given to you in class?

original post?that helicopter problem?

 

[cjl]

Yes, that one.

 

[gracy]

Yes, that one.

yes,it is part of my physics practice worksheet.but it is not about vertical free fall projectile.

 

[cjl]

It is, however, about a projectile in free fall from some initial, nonzero height, which means you must have some other form of the equations of motion (since your current version does not apply in that situation). It should probably look somewhat similar to the form given to you by BiGyElLoW above.

 

[BiGyElLoWhAt]

equation for Range =u square sin2theta/g
time of flight=2u sin theta/g
maximum height=u square sin square theta/2g
here u=initial velocity
what i am missing?
is u not equal to initial velocity.then u is what ?

Sorry I meant to add this quote to that post.

i am sorry but i didn't understand your post.

What don't you understand? I can't help you if you're not more specific.

 

[gracy]

Sorry I meant to add this quote to that post.What don't you understand? I can't help you if you're not more specific.

that's what i didn't understand,as you had not added that quote.I want to ask a question,what would be maximum height in vertical free fall projectile,as it is going downward,maximum height would be height from where it is dropped?

 

[BiGyElLoWhAt]

Intuitively, yes. Plug in the values to the projectile motion equation i gave you (except use y) solve for the maximum.

 

[gracy]

Intuitively, yes. Plug in the values to the projectile motion equation i gave you (except use y) solve for the maximum.

let h=10 meters=height from where ball is dropped.and initial velocity=0 (both in x and y)I choose upward direction to be positive.and g=10 m/s square now
height at any given time t=10+0 - 5( t square)so that answer would always be less than 10.

 

[BiGyElLoWhAt]

Yes it will, which means mathematically (as well as intuitively), the initial height for an object being released from rest (or at least 0 vertical velocity) is the maximum.

 

[gracy]

Yes it will, which means mathematically (as well as intuitively), the initial height for an object being released from rest (or at least 0 vertical velocity) is the maximum.

what would be example of object having some vertical velocity when dropped from some height ?i can't imagine one.actually i want to ask what is initial velocity in projectile?is it the velocity with which something is dropped or thrown?but something is dropped or thrown with some force not with velocity.then what is initial velocity in projectile

 

[BiGyElLoWhAt]

An example would be if I had a ball in my hand while riding a skateboard, I went off of a ramp and was traveling upwards, then I simply let go.

You throw an object by exerting a force on the object over a time interval, which gives the object momentum, which is equal to mv. So you have given the object a velocity by exerting a force.

 

[gracy]

An example would be if I had a ball in my hand while riding a skateboard, I went off of a ramp and was traveling upwards, then I simply let go.

You throw an object by exerting a force on the object over a time interval, which gives the object momentum, which is equal to mv. So you have given the object a velocity by exerting a force.

so you mean whenever force is applied for some time interval,it necessarily imparts momentum and in turn velocity ,this velocity is initial velocity.but i have a doubt how some objects have initial velocity zero?some force must have been present with which the object is thrown ,then why that force didn't give momentum and in turn velocity to object?

 

[Doc Al]

but i have a doubt how some objects have initial velocity zero?

Just drop something from rest, don't throw it.

 

[gracy]

Just drop something from rest, don't throw it.

hmmm...dropping something doesn't require force instead it is taking our force back which we were exerting by holding it.so in throwing case initial velocity can not be zero?

 

[MayCaesar]

hmmm...dropping something doesn't require force instead it is taking our force back which we were exerting by holding it.so in throwing case initial velocity can not be zero?

Recall the First Newton's Law. It says something about inertial systems. Try to apply it to this problem. What inertial system would you consider using in this particular case?
Hint: try to find a system in which velocity of the dropping package will be easy to determine.
Hint 2: assume that air resistance is negligible, zero. What does it imply?

 

[gracy]

Recall the First Newton's Law. It says something about inertial systems. Try to apply it to this problem. What inertial system would you consider using in this particular case?
Hint: try to find a system in which velocity of the dropping package will be easy to determine.
Hint 2: assume that air resistance is negligible, zero. What does it imply?

Newton's first law of motion -An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an external unbalanced force.how this can help me to get the answer of my question -in throwing case initial velocity can not be zero?

 

[MayCaesar]

Newton's first law of motion -An object at rest stays at rest and an object in motion stays in motion with the same speed and in the same direction unless acted upon by an external unbalanced force.how this can help me to get the answer of my question -in throwing case initial velocity can not be zero?

Suppose you are standing in your room holding an apple. You drop an apple on the floor by just opening your palm, without any additional motion. What is the initial vertical velocity of the apple? What is the initial horizontal velocity?
Now, suppose you are standing in a helicopter moving to the east at a constant velocity. You drop an apple on the floor inside helicopter. What are the initial vertical and horizontal velocities in your frame system (connected to the helicopter)?
Now let's consider our case: you drop the apple and it falls down to the ground. How are its horizontal and vertical velocities change over time in your frame system? And how is your frame system moving relative to the ground?

If you are confused about the way the object is dropped, I believe it is assumed that no one throws it with force, it is just held in hands and then starts falling down solely by the force of gravity.

 

[gracy]

Suppose you are standing in your room holding an apple. You drop an apple on the floor by just opening your palm, without any additional motion. What is the initial vertical velocity of the apple? What is the initial horizontal velocity?

zero.

 

[gracy]

Now, suppose you are standing in a helicopter moving to the east at a constant velocity. You drop an apple on the floor inside helicopter. What are the initial vertical and horizontal velocities in your frame system (connected to the helicopter)?

initial horizontal velocity=that constant velocity with which the helicopter was moving.initial vertical velocity=zero

 

[lep11]

zero.

yes, it starts from rest, right?

 

[lep11]

initial horizontal velocity=that constant velocity with which the helicopter was moving.initial vertical velocity=zero

Yes, in that case the initial vertical velocity is zero but the horizontal initial velocity relative to the helicopter is also zero. The initial horizontal velocity relative to the ground is v_ox. When the package is dropped from the helicopter, the package is no more tied to the coordinate system of the helicopter but instead it is tied to the coordinate system of the Earth and we assume the surface of the Earth to be completely level when the distance traveled is "very small".

 

[gracy]

dropping something doesn't require force instead it is taking our force back which we were exerting by holding it.so in dropping case initial velocity can be zero.but in throwing case,force must be present to throw something that force always give momentum and in turn velocity .so initial velocity can not be zero in throwing object?

 

[lep11]

dropping something doesn't require force instead it is taking our force back which we were exerting by holding

You don't have to throw it down to cause the ball to fall because the force of gravity causes the object to accelerate towards the Earth as you are no more providing the normal force to counter gravity.