Projectile Motion HELP! (1 Viewer)

[cholly]

Is it possible to do the following question without the use of the
R= [u2sin (2x)]/g formula...ive never even seen it before but the success on e book use it :S

a ball leaves a players hand with the velocity of 12ms and 50 degrees to horizontal, travelling directly towards the baseline 18m away. would the ball land in the court?
vertical range is 4.3m.
{reference sucess one HSC space questions 3b]

 

[mattchan]

Yes it is possible

1) Use trigonometry to find Ux and Uy
2) Find Time by using v = u + at
3) Find range by using X = Ux x T


hope that makes sense. The range ends up being 14.5 i think...

 

[rama_v]

Yes, you can do it. Just find the horizontal component (its 7.71 ms) and then find the time it takes for the ball to drop from 4.3 m using the formula s=ut + .5 at^2 (you have to then double this value by two). After that simpy multiply the time by the value for Ux adn you should get 14.49 m which is stil in teh court

The question is a bit ambiguous in that it doest say what height the ball was thrown from so Im presuming is thrown frm the ground.

Damn Chan u beat me to it!

 

[cholly]

yay! thanks so much guys
btw...does anyone know what that otherformula is all about anyway?

 

[BlackJack]

It's combines the two steps, i.e. finding the time and calculating the distance, into one step.

x=ucosθ t
v_y=u_y-gt
t=2.u_y/g=2.usinθ/g.
.'. x = 2u²cosθ.sinθ/g = u²sin(2θ )/g.

 

[gordo]

i screwed up the projectile motion question in the hsc
it was like question 1 of section 2
i couldn't beleive it, i thought it was always so easy

my advice, practise emmmmmmmmmmm up to the day