Question

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

Answer 1

do you know general equation of a projectile?

Answer 2

I'll derive it if you want :(

Answer 3

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

Answer 4

@Callisto

Range*g * sin 2 θ = V^2

Answer 5

range = ?

Answer 6

okay what do you want now :/

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

u = initial velocity

Answer 11

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.

Answer 12

V^2=Vx^2+Vy^2

Answer 13

You should explain why dvy / dt = g then.

Answer 14

Okay so,