help please!!! motion cambridge (1 Viewer)

[macochris1]

can somebody please do question 7 of Exercise 3G in the Cambridge 3u y12 book.
i seem to me missing something pretty obvious, as i cant even derive the 6 equations. most parts if possible

thanks!!

 

[khorne]

For the benefit of anyone who doesn't have the book: the question is:

A plane is flying horizontally at 363.6KM/h and its altitude is 600 metres. It is to drop a food parcel onto a large cross marked on the ground in a remote area.
a) convert the speed of the plane into metres per second
b) derive expressions for the horizontal and vertical components of the food parcel's displacement from the point where it was dropped (take g = 10m/s)
c) show that the food parcel will be in the air for 2*sqrt{30} seconds
d) Find the speed and angle at which the food parcel will hit the ground.
e) At what horizontal distance from the cross, correct to the nearest metre, should the plane drop the food parcel.

 

[macochris1]

ahhh yeh that would help thanks

lol anyone???

 

[GUSSSSSSSSSSSSS]

a) 101 m/s

b) vertical:
g = x(double dot) = -10
= dv/dt = -10
dv = -10dt
v = -10t + c
initial velocity is 0
v = -10t = dy/dt
dy = -10t.dt
y = -5t^2 + c
initial displacement = 600
c = 600
y = -5t^2 + 600

HORIZONTAL:
velocity = 101 = dx/dt
dx = 101dt
x = 101t + c
initial displacement = 0
x = 101t

c) IT WILL HIT THE GROUND WHEN y = 0
therefore: y = -5t^2 + 600 = 0
5t^2 = 600
t^2 = 120
t = sqrt120 = 2sqrt30

lol sorry i gotta go, but i'll finish off the solution later, someone else will probably post it up in the meantime anyway

 

[macochris1]

hmm, i understand all that, but why is it for some questions (like this one) inital velocity = 0. i was taught the little triangle thing with x(dot) = v.cos.(alpha). so for other questions, say: a stone is projected from a point off the ground, to get x(dot) and the rest, you use the given angle and initial velocity, into the triangle to get x(dot) and y(dot) when t = 0. but in this question, since you are not given enough information, PLUS its being dropped from a plane (different to being projected from the ground), how can u assume initial velocity = 0.

yeah thanks the rest is fine from then onwards, its mainly the concept.
cheers

 

[Drongoski]

hmm, i understand all that, but why is it for some questions (like this one) inital velocity = 0. i was taught the little triangle thing with x(dot) = v.cos.(alpha). so for other questions, say: a stone is projected from a point off the ground, to get x(dot) and the rest, you use the given angle and initial velocity, into the triangle to get x(dot) and y(dot) when t = 0. but in this question, since you are not given enough information, PLUS its being dropped from a plane (different to being projected from the ground), how can u assume initial velocity = 0.

yeah thanks the rest is fine from then onwards, its mainly the concept.
cheers

As it is dropped from plane:

1) horiz component of velocity = 101 and stays constant since it is assumed there is nothing to change its velocity .. i.e no (horizontal) force or resistance acting on it; that is to say horiz acceleration , whether helping or going against it (negative acceleration) = 0

2) vertical component of velocity = 0 initially (i.e 0 sec from when it was dropped). There is a force working on it, called gravity, and it causes a downward acceleration of 10 m/sec.

3) so as time t sec progresses, horiz velocity = 101 (constant) whereas the vert velocity increases downwards.

4) the resultant (or 'final' or 'net' or 'overall') velocity at any time t secs is determined by the horiz & vert velocity components. u can use a right-angled triangle to work this out (both the magnitude, u call 'speed', and the angle against the horizontal).

Hope this long-winded explanation helps.

 

[GUSSSSSSSSSSSSS]

wow i forgot about this
i'll finish my answer xD:

d) we just established that it hits the ground when t = 2sqrt30
therfore y(dot) = 5(2sqrt30)^2 + 600 = 1200
AND x(dot) = 101
so by drawing that triangle thingy, with one side = 101, the other = 1200
velocity = sqrt(1200^2 + 101^2) = *lol sorry i dun have a calculator on me* (by pythagoras' theorem)
and the angle is tan(inverse) (1200/101)

e) the distance between when it was dropped (x = 0) and where it lands is the answer
so: x = 101t
we know that it hits the ground when t = 2sqrt30
so x = 101*2sqrt30
(once again i dun have a calculator sorry)


HOPE IT HELPED xD, and hopefully i didn make a mistake, im sorta tired =P

 

[addikaye03]

m/s=km/3.6h, basically 3.6 is the magic number that i always remember, its just 60minx60secs/1000km... the rest of the Q has already been answered in much detail by other BOS members

 

[macochris1]

haha wow thanks everyone.

you explained it soo much better than my tutor LLOL.
cheers