Question

@satellite73 Equation of trajectory

Answer 1

The equation of a projectile is \[\huge y=\sqrt{3}x - \frac{ gx ^{2} }{ 2 }\]
Find the angle of projection. Also find the speed of projection.
Where t=o and y=0
also \[\huge \frac{ d ^{2}x }{ dt ^{2} }=0\]
and \[\huge \frac{ d ^{2}y }{ dt ^{2} }=0\]

Answer 2

@Miracrown

Answer 3

@iambatman

Answer 4

Okay. I will try this :-)

Answer 5

Yes go ahead

Answer 6

We need to write down the equation of trajectory first.

@No.name can you write it here? We will compare that with the equation given.

Answer 7

There are 6 forms of theequation of trajectory

Answer 8

We will use this form here :

\(y = x\tan \theta - \cfrac{gx^2}{2v^2 \cos^2 \theta}\)

Answer 9

Okay

Answer 10

so tan theta is root 3

Answer 11

Yeah.. I think so. Do you have any options?

Answer 12

Yes it is 60

Answer 13

So finding out speed is easy then

Answer 14

I didn't get the calculus part

Answer 15

Yes, fine. So, we have theta = 60 degrees.

Now, \(2v^2 \cos^2 \theta = 2\)

\(v^2 \times \cfrac{1}{4} = 1\)
\(v^2 = 4\)
\(v = \pm 2\)

Answer 16

Same here. I'm trying to understand the calculus part.

Answer 17

Is v = +/- 2 the answer of the question for speed?

Answer 18

+2

Answer 19

Yeah right!

Answer 20

I don't get why the calculus part is given to us here. There is no need for that!

Answer 21

me too

Answer 22

Thanks a million, you learnt projectile motion?

Answer 23

\(x = vt \cos \theta\)
\(\cfrac{dx}{dt} = v\cos theta \times 1\)
\(\cfrac{d^2 x}{dt^2 } = v\cos \theta \times 0 \)
= 0

Yes, I learnt it earlier.

Answer 24

Okay, but there are no answers given for the calculus part

Answer 25

May be they wanted us to find that ?

Answer 26

yeah maybe , thanlks!!

Answer 27

You're welcome :-)

Answer 28

I don't understand, please explain me
from the given equation, it's a downward parabola with the vertex almost (0,0), how can it be a trajectory?
if we consider the equation of trajectory, then d^2y / dt^2 = a_y =9.8, it can't be =0 as given information.
too many unclear information to me. Please , make it clear. :) @No.name @mathslover

Answer 29

@vishweshshrimali5 will help us out.

Answer 30

@Loser66 \(vt\cos \theta\) is the x-component in the projectile. The acceleration of the x-component is zero, so mathslover is not wrong.

Answer 31

I am talking about y , not x!! and \(\dfrac{d^2y}{dt^2}=0\) how?

Answer 32

Don't worry, I'm getting \(\frac{d^2 y}{dx^2} = g\).

Answer 33

Whoops, didn't consider the sign. I meant \(-g\).

Answer 34

One more thing!! I didn't see the work of @mathslover . I was too busy to understand what is wrong with the problem and why we can't use calculus part to solve. I ended up with my questions

Answer 35

Wait wait wait..

it is \(\cfrac{d^2y}{\color{blue}{\bf{dt^2}}}\) and NOT \(\cfrac{d^2 y}{\color{blue}{\bf{dx^2}}}\)

Answer 36

That's why : \(\cfrac{d^2 y}{dt^2} = 0 \) ...

Answer 37

yes, @mathslover
I was taught that for this kind of problem, I have to write 2 equations for x and y. In other words, I have 2 parametric equation for x, y w.r.t time t.
x = x_0 + v_(0x)t + 1/2 a_xt^2 , and solve for t to plug back to equation y = y_0 +v_(0y) +1/2a_yt^2
and from the given information d^2x/dt^2 = a_x =0. It's perfectly ok
but d^2y/dt^2 =0 is so nonsense

Answer 38

\(y = v \sin \theta t - \cfrac{1}{2} gt^2 \)
\(\cfrac{dy^2}{dt^2} = -g \)
hmm yeah @ParthKohli was right with the RHS.

Answer 39

oops d^2y *

Answer 40

\[y = vt\sin(\theta) - \dfrac{1}{2}gt^2\]\[y' = v\sin(\theta) - g t\]\[y' ' = -g\]And\[y = x\tan(\theta) - g x^2\dfrac{1}{2v^2 \cos^2 \theta}\]\[y'= \tan(\theta) - g \frac{1}{v^2 \cos^2 \theta } \]\[y'' = 0\]So we basically intermixed our results. lol

Answer 41

it should be that , but he, the Asker, said \(\dfrac{d^2y}{dt^2}=0\) ahaaaah!! he drove me crazy

Answer 42

Whoops.

Answer 43

both y are different .. one is equation of trajectory and one is for vertical direction.

Answer 44

Its getting confusing. Wait, lemme check my book :-)

Answer 45

Hmm, both of them are the same. One is a function of \(t\) and another is a function of \(x\), so as to calculate the second derivatives in terms of both.

Answer 46

You see, if we ignore the t, just focus on the given equation, it has no \(y_0\). Does it mean the projectile starts from the ground where y_0 =0, right? but the vertex ( maximum) height is (0,0) hahahaha....

Answer 47

I think its wrong..! (the statement)

\(\cfrac{d^2 y}{dt^2} \ne 0\) ... as acceleration along y axis can not be sero.

Answer 48

*zero

Answer 49

\(\cfrac{d^2 x}{dt^2} = 0 \) that is acceleration along x axis is zero. But, if \(\cfrac{d^2 y}{dt^2} = 0 \) then it will conclude that acceleration along x axis and acceleration along y axis is zero.

@Loser66 is right. I don't think that the question is right. As, the vertex is zero and the initial position is also zero.

Answer 50

Oh.

Answer 51

Sorry, I meant the *velocity* is constant. And I read that as d2y/dx2 (so many mistakes argh!).

Anyway, we've already shown why d2y/dt2 is -g.

Answer 52

Well, IF \(\cfrac{d^2y}{dt^2} = 0\) then \( y = \sqrt{3} x\) which is a straight line.

Answer 53

ok, I satisfy with you guys conclusion. At least, my doubt are off. hihihihi

Answer 54

:-)

Answer 55

The statement is correct , i verified it with my prof.

Answer 56

@No.name - Professor verified it? Its impossible bro.

See, if you are given with \(\cfrac{d^2y}{dt^2} = 0\) and \(\cfrac{d^2 x }{dt^2 } = -0\) then :

For \(\cfrac{d^2y}{dt^2 } = -g = 0 \) and for \(\cfrac{d^2x}{dt^2}= a_x = 0\)

Thus, note that -g = 0 or g = 0

The given equation of trajectory becomes :

\(y = \sqrt{3} x \) which is an equation of straight line.