Question

for angle of projection (45-theta) degree and (45+theta) with the horizontal of a projectile launched upwards from leve; ground if the range of the projectile are r1 and r2 for the same velocity then


r1=r2 r1=3r2/2

Answer 1

theta =35

Answer 2

how

Answer 3

range=(velocity * sin2 theta)/gravity

Answer 4

range=(velocity^2 * sin2 theta)/gravity

Answer 5

ok

Answer 6

the answer is r1=r2

Answer 7

v1=v2 ... two conditions cnt be true ...r1=3r2/2

Answer 8

no the answer is r1=r2

Answer 9

how

Answer 10

me too confused///

Answer 11

r1=r2 r1=3r2/2 ... your qustn contains both these conditions ... i took v1=v2 and r1=3r2/2

Answer 12

r1=r2 r1=3r2/2 these are optionss

Answer 13

ooooh those r d ans :P ... i thought those r conditions.. let me solve n get u bak

Answer 14

r1=v0 cos (45- theta) t1
t1 = 2v0sin(45-theta)
r1=v0^2sin(90-2 theta)
r2=v0 cos (45 + theta) t2
t2 = 2v0sin(45+theta)
r2=v0^2sin(90+2 theta)
r1/r2=sin(90-2theta)/sin(90+2theta)=1
since
sin(90-alpha)=sin(90+alpha)
r1=r2

Answer 15

I hope you know that
sin(2A) = 2 sin A cos A
I used this between the second and the third step

Answer 16

If you have any question feel free to ask.

Answer 17

how t1 = 2v0sin(45-theta) ??

Answer 18

can we use R=u^2sin 2 theta????? to get this

Answer 19

plzz try to that in that way plzzz

Answer 20

Yes you can use it except put at denominator g
R = (u^2sin 2 alpha)/g
Sorry I made a mistake in the formulas
r1=v0^2sin(90-2 theta)/g

Answer 21

ok

Answer 22

v=v0y-gt
t1 is the full time of the flight so when object is at the top t1/2 time passed
at the top v=0 so
0=v0y-g(t1/2)
t1=v0y/2=v0 sin(theta-45)/g
oops there is another mistake.
That's how i got it

Answer 23

Got it?

Answer 24

no...........

Answer 25

sorry
t1=2v0y/g=2v0 sin(theta-45)/g

Answer 26

Now?

Answer 27

no plzzz make it complete and clear plzzz

Answer 28

Look at it this way:
\[r1 = u^2 \sin 2(45- \theta)/g \rightarrow u^2 \sin(90- 2\theta)/g\]
\[r1 = u^2 \cos 2\theta /g\]

Answer 29

\[v=v_{0y}-gt_{1}\]
t1 is the full time of the flight so when object is at the top t1/2 time passed
Object first rises then falls, when it is at the top only half of the time is passed.
at the top v=0 so
\[0=v_{0y}-g(t_{1}/2)\]
\[t_{1}=2v_{0y}/g=2v_{0} \sin(\theta-45)/g\]

Answer 30

ok then r2??

Answer 31

similarly,
\[r2 = u^2 \sin 2(45 + \theta)/g\]
\[r2 = u^2 \sin(90+ 2\theta)/g\]
\[r2 = u^2 \cos2\theta/g\]

Answer 32

so r1 = r2

Answer 33

how sin(90 +tetha)=cos2theta??

Answer 34

sin( 90 + theta)= cos theta. its a trigonometric formula

Answer 35

sin(90-thetha)=cos tetha is it???

Answer 36

yes

Answer 37

then sin(90 +tetha)=costhetha possible???

Answer 38

yes. those are fundamental formulae.

Answer 39

no it is not
sin(90 + theta) = -cos theta

Answer 40

yea u r correct

Answer 41

OH NO I AM correct I am so sorry CHAND is right

Answer 42

you can check it sin 135 = cos 45

Answer 43

@shameer1

Answer 44

the method of chand is very essy to understand can ArchiePhysics do that way or chand continoue ur ans

Answer 45

my answer is completed. in archie's method, you get to understand the underlying physics of the projectile motion and my method is just answer oriented.

Answer 46

ok thanzzzz

Answer 47

Sometimes I do mistakes though they could be minor it is because type text uncarefully. Sorry!

Answer 48

never mind!

Answer 49

Thanks!

Answer 50

http://openstudy.com/study#/updates/4fabe009e4b059b524f78c10

Answer 51

ArchiePhysics u helped me lot in lot of situvation thanxxxx

Answer 52

You're welcome, and check my typos))), please!