Show that for a projectile d^2(v^2)/dt^2=2g^2

Question

aravindg · Answer

do you know general equation of a projectile?

hba · Answer

I'll derive it if you want :(

callisto · Answer

Ok, you need to derive it.
Derive v^2

hba · Answer

@Callisto 

Range*g * sin 2 θ = V^2

callisto · Answer

range = ?

hba · Answer

okay what do you want now :/

hba · Answer

If i knew everything i wouldn't have posted the ques.

callisto · Answer

You can derive the range too...
But...
My approach was to find $v_x$ and $v_y$
So, $v^2 = v_x +v_y$

callisto · Answer

$v_x$ and $v_y$ will be in terms of u, g and t
I think you can take u and g as constants.

callisto · Answer

u = initial velocity

hba · Answer

V2 =V; +V;. For a projectile Vx is constant, so we need only evaluate ~(V;)/dt2. The first
derivative is 2vy dvy / dt = -2vyg. The derivative of this ( the second derivative) is -2g dvy / dt = 2g2.

hba · Answer

V^2=Vx^2+Vy^2

callisto · Answer

You should explain why dvy / dt = g then.

hba · Answer

Okay so,

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