harder projectiles (1 Viewer)

[its_bui]

a stone is thrown so that it will hit a bird at hte top of a pole. However, at the instant the stone is thrown, the bird flies away in a horiztonal straight line at a speed of 10metres per second. The stone reaches a height double that of hte pole and, in its descent, touchs the bird. Find the horiztonal component of hte velocity of the stone.

ps. no joels or anthonys allowed

 

[Affinity]

this is from fitzpatrick's 3 unit book

 

[its_bui]

yea i know lol but i though it was a pretty good question, good enough to fall under harder 3u topic

 

[richz]

lol, nice question, sorry cant help... not up to this topic yet..
Wow ur already u to this???? What topics have u guys done???

 

[its_bui]

we havent done that topic in class yet , i do this stuff with my tutor or by myself

 

[dawso]

lol, yes, this is from fitzy's 3unit book, its an absolute bitch, a very famous question

i have a solution and even still i dont get it, id say just move on....
(or wait for ngai 2 come on)

 

[its_bui]

i have the solutions too and i don't get it either, but after 2 hours i finally got it!~

 

[Ogden_Nash]

It took me a while to get it but I'll quickly list the steps I took...

1) Use the fact that max height = 2h to find an expression for h in terms of V, @ and g.

2) Use the quadratic formula to find the 2 values of t when y = h.

3) Use x = Vtcos@ and sub the previous values of t to find the horizontal displacements when the projectile is at height h.

4) The difference of these values will give you the distance that the bird has travelled at 10ms^-1.

5) Use distance = time*speed for the bird's flight using the appropriate values you've calculated and you'll end up with a value for Vcos@, which is what you needed to find.

I ended up with Vcos@ = 12.07ms^-1

 

[ngai]

a stone is thrown so that it will hit a bird at hte top of a pole. However, at the instant the stone is thrown, the bird flies away in a horiztonal straight line at a speed of 10metres per second. The stone reaches a height double that of hte pole and, in its descent, touchs the bird. Find the horiztonal component of hte velocity of the stone.

oh this question...
there's a really elegant solution to this...

the stone goes up to 2H and then down to H, then hits the bird
let the time taken to fall a vertical distance of 2H be T - this is the same time taken to rise a vertical distance of 2H
also, let time taken to rise or fall H be U

vertically, y=gt²/2, so 2H = gT²/2 and H = gU²/2
dividing those two gives 2H/H = 2 = T²/U²
so T = (sqrt2) U

draw a picture, and call the horizontal distance from the bird's start to the bird's end D
bird travels D in T+U secs at 10 m/s, so D = 10(T+U) = 10(sqrt2+1) U
stone travels D in U+U secs at v m/s, so D = 2vU
2vU = 10(sqrt2+1) U
v = 5(sqrt2+1)
= 12.07

 

[its_bui]

hmm my solution was not as good as this but i was wondering
how you figured this out:

bird travels D in T+U secs at 10 m/s, so D = 10(T+U) = 10(sqrt2+1) U
stone travels D in U+U secs at v m/s, so D = 2vU

it doesn't make sense to me

 

[ngai]

hmm my solution was not as good as this but i was wondering
how you figured this out:

bird travels D in T+U secs at 10 m/s, so D = 10(T+U) = 10(sqrt2+1) U
stone travels D in U+U secs at v m/s, so D = 2vU

it doesn't make sense to me

the total time of everything is the time taken for the stone to go up 2H then down H, ie. total time = T + U
bird flies for the whole time to travel that horizontal distance D

for the stone to travel horizontal distance D, it takes less time..
in fact, the stone goes up H and then down H during the time it takes to travel D horizontally
so time for stone to travel D is U + U

then after that, its just distance = speed*time

 

[dawso]

watd i say.......wait 4 ngai