PCB Heaven
General Category => Maths, Geometry, Physics and others => Topic started by: piyt on September 01, 2008, 22:19:35 PM

I am a studet and i would really like your help of possible
Two boys play with a slingshot. There is a tree with height 14 meters. One boy say to other "I will throw this stone over the tree with the slingshot"
The other laught and say "no way"
The boy throw the stone with the slingshot. The stone left the slingshot with speed 90km/h.
Did it cross the tree from over?
Any help please this week
Thank you from now

Don;t you need the weight of the stone??????

Don;t you need the weight of the stone??????
I really do not know the answer.I only know that the weight might not be important. It would be important if you want to calculate how much power the sling shot had to gain this speed. But from the time it reaches this speed, the travel time and distance only depends on the angle (and friction). See the escape velocity (http://pcbheaven.com/forum/index.php/topic,117.0.html) post.

I also don't know the answer but i think something's missing to.

Well, i somehow made a mistake. I say : "Did it cross the tree from over?" but actually i translate wrong and is like :"CAN it cross the tree from over?". Meaning that is it possible?
And of course, the air friction is 0. But i suppose this need not to mention.

Well, then i should suppose that the boy throw the stone with 45 degrees angle... Let's see.
First, to help us in calculations, we convert 90km/h to m/s. This is:
90*1000/3600 = 25m/sec
We call X the horizontal travel, Y the vertical (and what we are looking for...) and V the velocity, Vi the initial and Vf the final, a the acceleration, t the travel time.
X=V_{IX}t+0.5 x a_{X} x t^{2} (1)
V_{fX}=V_{iX} + a_{X} x t (2)
V^{2}_{fX} = V^{2}_{iX} + 2a_{X} x X (3)
these are the formulas for our calculations for the horizontal motion. And we can say that a_{X}=0, as the stone gets no more acceleration. We rewrite the above as:
(1)=> X=V_{IX}t (4)
(2)=> V_{fX} = V_{iX} (5)
So comes for the Y motion:
Y=V_{iY} t = 0.5 x a_{Y} t^{2} (6)
V_{fY}=V_{iY} + a_{Y} x t (7)
V^{2}_{fY} = V^{2}_{iY} + 2a_{Y} x Y (8)
When the stone goes to higher point, the vertical velocity V_{FY} will be 0. so:
(7)=> 0 = V_{iY} + a_{Y} x t
and also, the time that the stone will be at the higher point, is half of the total travel time. So, the above equation for the travel from slingshot to max height will result to:
0 = 25ms^{1} Sin (45)  (9.8ms^{1} x t)/2 => ........ => t=3.607sec.
This will be half the travel time, when the stone is up the sky ;D
Now, we use the equation (5):
(4)=>X=V_{IX}t => X=25 x Cos (45) x t = ..... => X=63,763 meters.
This is the distance that the boy should stand away from the tree, because at this distance the stone will be at the highest point.
Now with (8)
(8)=> V^{2}_{fY} = V^{2}_{iY} + 2a_{Y} x Y =>
0 = (25 x Sin(45))^{2}  2 x 9.8 x Y => .... (solve for Y) ... => Y = 15.94 meters........
If the boy stands 63,763 meters away from the tree and throw the stone with an angle of 45 degrees with 90km/h initial speed, then the stone will go over this tree...
I must admit, i had many many years to deal with projectiles...

Kam thank you. Thank you very much. You made my day
I search the internet but there is not such a complete and forme solution of this problem
I will study the solution. Thank you. 8)

bravo kam

You see? Weight is never important!!! I had decide to go on a diet but, it is no more important!

You see? Weight is never important!!! I had decide to go on a diet but, it is no more important!
:D lol
of course it is not important, as long as your feet are capable to give you the speed with such a... aaaa... aa... a mass!