Question

A particle is projected in gravity such that its x and y cordinates are related s:
y=ax-bx^2
then speed of projection=?

Answer 1

@Yahoo! @ujjwal

Answer 2

@ghazi

Answer 3

I tried somethings..

Answer 4

\[\LARGE y=xtan \theta-\frac{gx^{2}}{2u^{2}\cos^{2} \theta} \]

Answer 5

\[\frac{ d }{ dx }(ax-bx^2)=(a-2bx) m/s\]

Answer 6

comparing..
a=tan theta and b = rest of the stuff

Answer 7

what next

Answer 8

i thought it would be just differentiation i didnt see its a projectile motion

Answer 9

its projectile ! :P
its "PROJECTED"

Answer 10

lol yea yea hold on

Answer 11

\[\LARGE x=\tan \theta(1-\frac{X}{R}) \]
if we compare with this..
\[\LARGE y=x(a-bx) \]
divide by a
\[\LARGE y=x(1-\frac{bx}{a})\]

Answer 12

R=tan theta?perhaps

Answer 13

when you have made the comparison i guess you can find velocity too , just from the second part , make comparison and your net velocity will be equal to \[V=\sqrt{V _{H}+V _{V}}\]

Answer 14

Options:
\[\LARGE \sqrt{\frac{g(a-1)}{2b}} \]
\[\LARGE \sqrt{\frac{g(a^{2}+1)}{2b}} \]
\[\LARGE \sqrt{\frac{g(a+1)}{2b}}\]
\[\LARGE \sqrt{\frac{g(a^{2}-1)}{2b}} \]

Answer 15

in your comparison you have forgotten g

Answer 16

1st one?

Answer 17

2nd one

Answer 18

where is G :S

Answer 19

\[\LARGE Y= x \tan \theta ({1-\frac{x}{R})}\]

Answer 20

\[\LARGE y=x(a-bx)\]

Answer 21

the thing that i am saying is \[\tan \theta=a\] and \[b=\frac{ g }{ 2u^2\cos^2 \theta }\] apply the two components of velocity and take out the resultant and i guess you will get option B

Answer 22

"apply the two components of velocity and take out the resultant "
can u be a bit precise?

Answer 23

hmm if you are concerned about precision , i am trying me best to explain though i know i am unable to explain but i am tryna hard actually what i am saying is to fine Vh and Vv and take resultant

Answer 24

HOW to get Vh and Vv :/
we dont hv anythng

Answer 25

@experimentX

Answer 26

find the vertex of that parabola. using that, find the horizontal component of velocity. and also find the time of flight.

then find the range of the parabola. using the time of flight, find the horizontal component of velocity.

using these two, find the speed.

Answer 27

complex

Answer 28

i dont even know how to find the vertex

Answer 29

there is nothing complicated about it. put x=0, you get y=0,
at vertex the slope is zero.

Answer 30

so dy/dx = 0 .. get the value of x, put that x into above and get the value of y.

Answer 31

i cant follow up..im not that advanced :/ not your fault..isn't there an easier method? thanks for helping though!

Answer 32

why cant we work out with compring

Answer 33

there are couple of ways of finding the vertex of parabola.