Projectile Motion (1 Viewer)

[ludac]

Talia is holding the garden hose at ground level and pointing it obliquely so that it sprays water in a parabolic path 2 meters high and 8 meters long. Find, using g= 10m/s^2, the initial speed and angle of elevation, and the time each droplet of water is in the air. Where is the latus rectum of the parabola

 

[random-1006]

Talia is holding the garden hose at ground level and pointing it obliquely so that it sprays water in a parabolic path 2 meters high and 8 meters long. Find, using g= 10m/s^2, the initial speed and angle of elevation, and the time each droplet of water is in the air. Where is the latus rectum of the parabola

doesnt seem that bad

we know max height is 2, and the range is 8

we know corresponding expressons for these as well:

let a be launch angle

now we have : V^2 sin(2a) / g =8
V^2 (sina)^2 / 2g =2
expanding sin2a as 2 sinacosa, and dividing eqn 1 by eqn 2

we get tana=1

rearranging eqn 2 for V^2 and subing into second half of eqn below ( stuff in bold below)



now the path ( y in terms of x)

y= xtana - [g (seca)^2 x^2] / [2V^2]
= x - [x^2 (tana)^2]/ 8, but tana=1

therefore y= x -x^2/ 8 { now complete square}

y= -(1/8) { x^2 -8x}
y= -(1/8) { x^2 -8x +16 -16} ( note the -16 was expanded out, and two negatives make a positive}
y= -(1/8) [ x-4]^2 +2
(x- 4)^2 =-8 ( y- 2)


a=2, and its concave down

so vertex= (4,2)
focus= (4, 0)
now latus rectum goes through focus, perpendicular to the axis of symmetry

latus rectum: y= 0 { ie the ground}

this should make sense as remember the length of the latus rectum= 4a { this is the "a" in x^2 = 4ay, NOT the launch angle} = 4(2)= 8= RANGE in this case, so you can be pretty sure you havent made a mistake.


the other parts of the question are easy ( we know that tana=1, ie a= 45 degrees)

from eqn 1:

V^2 sin(2a) / g =8 {you can find V as you know a= 45 degrees)}

then initial x velocity= Vcosa, initial y velocity= Vsina

and the initial velocity ( all together, velocity= sqrt ( (x velocity)^2 + (y velocity)^2 )

and time of flight is standard.

of course in an exam you would have to derive ALL THESE formulas ( eqn of path, time of flight, max height, range) , if this question came up it would be worth like probably 12 marks because there is so much work. really nice question

 

[cutemouse]

if this question came up it would be worth like probably 12 marks because there is so much work. really nice question

You mean, more like 6 or 7 marks?

 

[random-1006]

You mean, more like 6 or 7 marks?

ok 1 mark for each formula that you derive ( thats 4), 1 mark for getting tana=1, 1 mark for getting y= x -x^2/8 , 2 marks for getting eqn of latus rectum, then another 1 mark for finding initial speed, maybe another mark for the time of flight. there is a fair amount of working noob. ok about 9marks ( maybe 10), The point is, its a fair few. i reckon this is probably about 9 marks ( of Q5-6 level difficulty )

 

[cutemouse]

see i like questions like that which involve 2 unit knowledge, would catch out someone like you easy.

I think I pretty much burned you with that probability question

ok 1 mark for each formula that you derive ( thats 4)

If you're talking about the eqns of motion then you wont get 1 mark for each formula (ie. for x, dx/dt, y, dy/dt). They don't mark that way in the HSC generally.