Cambridge Projectile Motion (1 Viewer)

[dishab]

Hey can someone please help me with this question no. 17 about projectile motion.

 

[random-1006]

Hey can someone please help me with this question no. 17 about projectile motion.

im not putting up the working, i will give a good description on how to do it,

a i. we know x and y co ordinates are equal , solve Utcosa = Utsina - 1/2 g t^2

ii. most likely algebra, find time of flight ( standard, then do some manipulating)

iii. we reuqired t2 > t1, so 0< 1-cota < 1 [ note, if you divide by a number between 0 and 1, your result is LARGER, we cant divide by zero]

solve 1- cot a > 0 ( ie 1 - 1/ (tana) >0) { careful, you get an unknown in the denominator, multiply by the square of the denominator} and apply approriate arguments e.g. if you get an obtuse angle the ball would be being thrown the direction OPPOSITE that shown in the picture. and it obviously cannot be greater than pi/2 and cannot equal pi/2 or else it would go up and fall stay back don vertically


b. i. no acceleration in horizontal direction, constant x velocities!, result follows

ii.

Cos B = U cos (a) / V now a= 2b
Cos B = U cos (2B) / V

using double angle expansion for cos

you get it down to 2 ( cos B ) ^2 - (V cosB / U) -1 =0

use quadratic formula, simplify.

one tof the lines you get is

cos B = [ V/ U +- sqrt ( V^2/ U^2 +8) ] / 4

cos B must be positive as B= 1/2 a , and a is acute.

now, it should be easy to see that sqrt ( [V^2 / U^2] +8) will be larger than V/ U, hence we must take the positive sign the formula to make sure cos is positive, simplify a bit more and thats it.


those are some good questions, really testing everything!

 

[dishab]

I dont get ii from part a)

 

[random-1006]

I dont get ii from part a)

ok, now to show the result it would be equivalent to show that T1/ T2= 1-cota yes?? { im doing it this way as it is much easier to play around with ratios and cancel factors etc, than substituting and mulitplying out and showing that LHS= RHS }

ok now T2 is the time of flight, easy to derive , T2= (2Usina) / g

now T1 / T2= [ 2U/ g ( sina -cosa) ] / [ 2Usina / g ] { cancelling down}
T1/T2= (sina -cosa)/ (sina) { (a +b) / c = a/c + b/c }

= [ ( sina)/ (sina)] - [cosa/sina]
= 1 - cota

they are really good questions

 

[dishab]

Awww i see. Thnx random-1006

 

[random-1006]

Awww i see. Thnx random-1006

lol i remember looking at this question last year but i gave up on it, dont know why i can get it now, probably because i was stressed before the hsc when i saw it.