Projectile motion question (1 Viewer)

[acevipa]

A projectile is launched at an angle of 30° and reaches a range of 2.90 m in a total time of 0.57 seconds. Ignoring air resistance, determine the:

a) Maximum height
b) Velocity

Also, just a question off this, is it possible to determine the maximum velocity and if so, how?

I'm not too sure if I'm right, but I calculated 0.44 m for max. height. and got a velocity of 5.88 m/s.

But, I assume that is the initial velocity and that would change.

 

[Trebla]

Maximum height occurs when the vertical component of velocity is zero, because in terms of the vertical dimension it stops momentarily before falling back.

 

[cutemouse]

A projectile is launched at an angle of 30° and reaches a range of 2.90 m in a total time of 0.57 seconds. Ignoring air resistance, determine the:

a) Maximum height

v_y=u_y+a_yt
Where v_y=?
u_y=0
a_y=9.8
t=0.57/2

Then use v_y²=u_y²+2a_yΔy, and find Δy

b) Velocity

I assume they mean initial velocity?

v=sqrt(u_y²+v_y²) 30 degrees below horizontal.

 

[cutemouse]

The projectile never stops in the air. It goes through zero, but never actually stops.

Maximum height occurs when the vertical component of velocity is zero, because in terms of the vertical dimension it stops momentarily before falling back.

 

[Trebla]

v_y=u_y+a_yt
Where v_y=?
u_y=0
a_y=9.8
t=0.57/2

Then use v_y²=u_y²+2a_yΔy, and find Δy

It should be v_y = 0, not u_y. If u_y = 0, it would suggest the projectile started with an angle of inclination of 0. Also, a_y = - 9.8 because by convention vectors directed upwards are positive.

The projectile never stops in the air. It goes through zero, but never actually stops.

I think I should have put the word stop in inverted commas. It theoretically does "stop" since it passes through v_y = 0, but for an extremely small increment of time.

 

[cutemouse]

It should be v_y = 0, not u_y. If u_y = 0, it would suggest the projectile started with an angle of inclination of 0. Also, a_y = - 9.8 because by convention vectors directed upwards are positive.

Well, if you think about it, it's the same as the projectile being dropped from a height, instead of it starting from the ground and thus Uy=0, but yeah your method probably is better.

I think I should have put the word stop in inverted commas. It theoretically does "stop" since it passes through v_y = 0, but for an extremely small increment of time.

Theoretically it doesn't stop. Practically, I don't know as it does depend on air resistance and all these factors.

Still think that it stops? Well then why don't you tell me exactly how long this projectile (op's question) 'stops' in the air for?

 

[tanguyen]

Wait are you being serious jm01? it MUST stop.
If you have a number line and draw a line from -2 to +2, you MUST go through zero. The same applies with velocity (where the + and - signs are now direction).
But you don't say "how long" it stops in the air for, its termed "instantaneous". You sorta can't think about this as if things are travelling at constant speed; everything (velocity in particular) is changing in every moment.
You could start delving into Calculus and differentiation but that should be sufficient.

If you were joking I just sounded like a major douche.

 

[cutemouse]

Yes, it goes THROUGH zero; But it never actually stops in the air

 

[cutemouse]

Also, I think you should consider the phrase "pot calling the kettle black". You're being so god damn serious.

I was just trying to make a point that, unless if you can tell me how long the ball or projectile 'stops' in the air for, then it doesn't stop in the air.

 

[Trebla]

It "stops" in terms of vertical dimensions for an infinitesimal time interval of dt = lim _{Δt --> 0} Δt
where Δt is an increment of time ...lol

 

[kaz1]

Yes, it goes THROUGH zero; But it never actually stops in the air

The vertical component is 0 for a moment but the horizontal component stays the same throughout the time it is in the air so it does not actually stop.

 

[cutemouse]

No the vertical component never stops, its acceleration is 9.8ms^-2 throughout the flight (assuming that of course the vertical displacement is relatively small). It goes through zero, and yes at top of flight Vy=0, but it doesn't stop!

I'll repeat it again, if you're saying that projectile stops at Vy=0, then you need to tell me how long exactly it stops for.

 

[Trebla]

I told you already. It "stops" at "exactly" the infinitesimal time interval it reaches maximum height. i.e. it's a limit approaching a time interval of zero.

Your definition of "stop" seems to be zero velocity at a finite time.

My definition of "stop" includes the zero velocity at infinitely small time intervals which is analagous to the finite time interval approaching zero.

 

[cutemouse]

I told you already. It "stops" at "exactly" the infinitesimal time interval it reaches maximum height. i.e. it's a limit approaching a time interval of zero.

How long does the projectile stop for then?

 

[shaon0]

How long does the projectile stop for then?

an infinitely small amount of time.

 

[kaz1]

No the vertical component never stops, its acceleration is 9.8ms^-2 throughout the flight (assuming that of course the vertical displacement is relatively small). It goes through zero, and yes at top of flight Vy=0, but it doesn't stop!

I'll repeat it again, if you're saying that projectile stops at Vy=0, then you need to tell me how long exactly it stops for.

The acceleration is -9.8ms^-2 while it is going up so the velocity is decreasing but when it reaches the maximum height V_y=0 (infinetely small amount of time) then the acceleration becomes 9.8ms^-2 and the velocity starts increasing.

 

[Trebla]

How long does the projectile stop for then?

I already answered that question...twice...shaon0 and kaz1 answered it as well lol. You're not giving a counterargument against this answer.
Do you understand the concept of limits or infinitesimals? lol

 

[cutemouse]

Hahaha, I think this is like arguing that the aether exists....

Okay, an "indefinitely small time" -- Which is how long? How about giving me a figure?

Consider a projectile has been projected upwards. At t=5, the ball still has an upward velocity of v, how long does the projectile remain at v for?

Lets say t=10 is top of flight.

At t=15, when the projectile's velocity is downwards, this is the same point as t=5. How long does the projectile remain at v for?

Now, if the projectile does remain/stop at v in both those cases, and then adding the time the projectiles for in both those cases should be equal to the time that the ball stopped at the top of flight.

You can see that the balls don't stop at t=5 or t=15, so why should it at top of flight?

The acceleration is -9.8ms^-2 while it is going up so the velocity is decreasing but when it reaches the maximum height V_y=0 (infinetely small amount of time) then the acceleration becomes 9.8ms^-2 and the velocity starts increasing.

My point was that it is constant acceleration.

 

[youngminii]

jm I don't think you understand the whole concept of this 'infinitesimallly small' business.
Think about a simple number line from say, 0 to 1.
How many points/numbers are there between 0 and 1? An 'infinite' amount. Then what's the distance between the 'infinite' amounts of points/numbers? An infinitesimally small amount.

Same can be applied to a graph/parabola, which is pretty much the trajectory of a projectile.

In fact, think about the turning point in an upside-down parabola. That point is pretty much the 'stop' point of the projectile. How long is this 'stop'? An infinitesimally small amount

 

[W3S]

bla bla bla bla bla..... the vertical component never remains constant unless its sittin on the ground where gravity and earths resistance equal nil, however Horizontal velocity always remains the constant.
excuse me i've forgoten my text book but a few weeks ago my teacher tried doing a question where the cannon ball is fired from point a to b ignoring different gravity at higher altitudes and wind resistance and he got it wrong because he tried doing the whole parabolic curve. if u get a question where it says to find distance traveled use half the time eg only have one side with relevant information.