How Far Should a Helicopter Drop a Food Packet for Flood Victims?

[BiGyElLoWhAt]

equations of motions?suvat equations,right?we will not take projectile specific equations such as for maximum height,range,time of flight in this type of vertical free fall projectile motion?

Do it all if you want. It all works. In a vertical free fall (starting from rest) y_0=0 and x_0=0. Plug it in. I think it would be good to work out.

Edit: ^ That response time, though.:D

 

[cjl]

Edit: ^ That response time, though.:D

:D

 

[gracy]

starting from rest

with zero initial horizontal velocity)

with zero initial horizontal velocity or starting from rest i.e both vertical as well as horizontal velocity should be taken as zero for applying projectile motion equations to vertical free fall projectile motion?

 

[cjl]

You can do either. You can have the object start from rest at some initial height h₀, or you can have it start with some initial height h₀ and some initial vertical velocity V_y,0. Either way, you should get purely vertical motion.

 

[gracy]

You can have the object start from rest

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.

 

[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.