Projectile motion and gravitational acceleration

[Idoubt]

Homework Statement

A projectile is fired from the origin O at an angle 45 degrees from the horizontal. At the
highest point P of its trajectory, determine the radial and transverse components of it's acceleration in terms of the gravitational acceleration g

Homework Equations

x= vcos45 t
y= vsin45 t - 1\2gt²

The Attempt at a Solution

No clue

 

[cupid.callin]

how do you think will be the motion of projectile at max height?
parabolic, straight line, elliptical or something else?

 

[Idoubt]

I would say it is parabolic throughout the motion

 

[cupid.callin]

what is direction of velocity at top most point?

 

[Idoubt]

Horizontal.

 

[cupid.callin]

so now can you imagine that the instantaneous center of this velocity will be directly below it at some height.??
this center may or may not be on the ground

 

[Idoubt]

By centre you mean the centre of curvature right? yes I can see that, also if I know what that distance is, i can calculate the radial acceleration. But how do I find that the distance ( radius of curvature right? )

 

[cupid.callin]

yes i mean center of curvature ... happy you know that
last time i had to explain it to someone ... it took me 15 mins

So now, radial acceleration is along radius ...
here ... radius is vertical(right?) ... so what is vertical acceleration?

 

[Idoubt]

well the vertical acceleration is g, but then wouldn't that imply that the instantaneous radial acceleration is g itself?

 

[cupid.callin]

Yes that is correct!

 

[Idoubt]

well that seems great but the question was a multiple choice one and the choices were


1, a_r = 2g/(5)^1/2 ; a_theta= g/(5)^1\2

2, a_r = -2g/(5)^1/2 ; a_theta= -g/(5)^1\2

3, a_r = g/(5)^1/2 ; a_theta= 2g/(5)^1\2

4, a_r = -g/(5)^1/2 ; a_theta= -2g/(5)^1\2

 

[cupid.callin]

This doesn't make any sense

is the question correct? and also is the question of good level so that we can be sure that question is wrong?

 

[Idoubt]

Well, it was asked in a national entrance ( I'm from India ) for pg, for Indian Institute of technologies, which is the best technical institute in the country, I think it's probably safe to say the question is legit

here is a screen shot if i missed something.

RL=http://img220.imageshack.us/i/jamq.jpg/][PLAIN]http://img220.imageshack.us/img220/1197/jamq.jpg

Uploaded with ImageShack.us

 

[cupid.callin]

I'm also from india

is that an old question? if that is so there is a small chance that the question is wrong

can you tell me the year so that i can look it up in a book

 

[Idoubt]

jam ph 2007 question 10 . I downloaded the previous paper from their official site.

 

[Idoubt]

I figured it out now.

The prob was our origin. g is the radial acceleration only when the origin is the centre of curvature.

http://img691.imageshack.us/img691/1197/jamq.jpg

Uploaded with ImageShack.us

here a_r= gsin alpha

and a_theta= gcos alpha

the time for the whole flight T = (2)^1/2V/g ( by putting y = 0 in the quad eq )

and the time to reach max height will have to be half this since x is linear in time . so T_1/2 is v/(2)^1/2g

solving for y and x we get, y = v²/4g and x = v²/2g


alpha = tan^-1 (y/x) = tan^-1(1/2) = 26.565

then turns out, a_r = g/(5)^1/2 ; a_theta = 2g/(5)^1/2

but this is magnitude, I think both acc has to be negative since a_r is in the dir of decreasing r and a_theta is in the direction of decreasing theta, so the right answer should be D, am I right?