A particle is projected at an angle such that range and MAx height are same when again projected at same velocity but at different angle range is unchanged what is the ratio of the two time of flight?

Question

maheshmeghwal9 · Answer

Range & Maximum height are always equal only & only when body is thrown approximately at an angle of 76 degrees 
then i think ration is 1.

but btw t is the answer?

maheshmeghwal9 · Answer

t = wt*

anonymous · Answer

The options are   (a)  2      (b) 3     (c) 2.5

maheshmeghwal9 · Answer

wt is the different angle?

anonymous · Answer

since it is said that Range is same  if one angle is x  and other one is 90-x

maheshmeghwal9 · Answer

but u haven't stated in the question that range is unchanged @Yahoo! :/

maheshmeghwal9 · Answer

but where is this written that

range is unchanged?????????????

anonymous · Answer

sorry for the mistake

maheshmeghwal9 · Answer

np:)

anonymous · Answer

nw u read the question

maheshmeghwal9 · Answer

ok!
now it is much better:-)

anonymous · Answer

Range, Initial Velocity are same for both the projectiles but the angle is different? @Yahoo!

anonymous · Answer

Yup ....but that is second case

anonymous · Answer

There are two cases?! What's the first case?

anonymous · Answer

Range = Maximum Height ------> 1

anonymous · Answer

Oh

anonymous · Answer

but for same projectile right? and i have to determine the launch angle?

anonymous · Answer

we need the ratio of the two time of flight?

anonymous · Answer

okay, i get the question i think. thanks.

maheshmeghwal9 · Answer

In first case $$\frac{u^2 \sin ^2 \theta }{2g}= \frac{u^2 \sin 2 \theta }{g}$$

by solving we get $$\sin \theta = 4 \cos \theta.$$

so in this case time of flight = $$\frac{2 u \sin \theta }{g}=\frac{2u \space  4 \cos \theta }{g}.$$

In the second case 
time of flight = $$\frac{2u \sin {(90 - \theta)}}{g}= \frac{2u \cos \theta}{g.}$$

So we get ratio = 4.

maheshmeghwal9 · Answer

but this is nt the answer.
so this is also a problem.

anonymous · Answer

the answer should be 2

maheshmeghwal9 · Answer

I m also rechecking :|

anonymous · Answer

@maheshmeghwal9  hw the angle become equal in case 1

anonymous · Answer

and initila velocity

anonymous · Answer

there are diff .....i think

maheshmeghwal9 · Answer

no they aren't different
question is saying that
in first case when body is projected with initial velocity u & angle theta then 
ur range = max height

ok 
after this in second case also range & initial velocity is same but angle is different therefore angle is 90 - theta.

maheshmeghwal9 · Answer

gt it?

maheshmeghwal9 · Answer

gt it or nt @Yahoo! or i misunderstood the question?

maheshmeghwal9 · Answer

but why i m getting 4 as answer:/

anonymous · Answer

@Ishaan94  any suggestions

maheshmeghwal9 · Answer

@Ishaan94 sir plz help:)
@mathslover :D

mathslover · Answer

not so good .. :(

maheshmeghwal9 · Answer

np:)
thanx for replying:)

anonymous · Answer

if range and height are same, sin^2 x = 4 sin x cos x => tan x = 4. 

t1 = 2u sin x/g and t2 = 2u sin y/g

t2/t1 = siny/sinx, right?

maheshmeghwal9 · Answer

ya but y = 90 -x

anonymous · Answer

x and y are the launch angles for first projectile and second projectile.

anonymous · Answer

y isn't 90-x?

if range1 = range2 => u^2sin2x/g = u^2sin2y/g => sin2x = sin2. why isn't the answer 1?

anonymous · Answer

sin2x = sin2y*

maheshmeghwal9 · Answer

no range is same thefore
angles are complementary.

maheshmeghwal9 · Answer

& sin x = 4 cos x

maheshmeghwal9 · Answer

sin y = cos x

maheshmeghwal9 · Answer

sorry i gt 4 wrong


Answer is 0.25

100% sure
I think options have some mistake.:/

anonymous · Answer

can anyone suggest some gud video's on the topic motion in 2-D

maheshmeghwal9 · Answer

dude the solution is given by me
consult ur teacher 
my gr8 advice
answer is 0.25.

anonymous · Answer

projectile 1, speed initial = u, launch angle = x. we know x = arctan(4). 

projectile 2, speed initial = u, launch angle = y.

if range1 = range2$$\frac{u^2\sin 2x}g =\frac{u^2\sin 2y}g \implies \sin 2x =\sin 2y \implies x=y\quad or \quad 180 - 2x = 2y$$

ratio of time of flight is sinx/siny = 4

hmm seems like i am losing the grip

maheshmeghwal9 · Answer

no 
sin y / sin x = 1/4 = 0.25
will be answer .

anonymous · Answer

.25 is same as 4.

maheshmeghwal9 · Answer

ya u r right.

maheshmeghwal9 · Answer

but @ yahoo
options r wrong:/

anonymous · Answer

i dont think so...

maheshmeghwal9 · Answer

final conclusion
becoz we both gt same answers.

maheshmeghwal9 · Answer

hw can u say with such believe?

maheshmeghwal9 · Answer

I & @Ishaan94 both solved & gt 4 or 0.25
which one u take
both are correct

maheshmeghwal9 · Answer

i think the third option is 0.25 nt 2.5

maheshmeghwal9 · Answer

videos on 2d @ http://www.khanacademy.org/

maheshmeghwal9 · Answer

find them:)