Projectile Motion - formula derivation

[arpita45]

Homework Statement

present a formal derivation of the formula (given above) that was used to calculate the kinetic energy of the arrow immediately after release from the bow.. repeated here:

K = ½ x mgsH
Where m is mass, g is grav. acceleration and sH is horizontal velocity.
My mistake was that I found the derivation of the clue and not the equation above.

Homework Equations

clue : The above formula is derived from v^2 x sin 2 theta/g.
This formula is used for horizontal displacement, giving us sH = v^2 sin 2 theta. The problem is how do I get rid of the component of velocity?

The Attempt at a Solution

I`m not sure if this is correct but here it goes...
If we make sin theta independent then - sin (theta) = 2v^2/g
Now bring g to the other side, we get, sin(theta)x g = 2v^2
Now to make v^2 independent by bringing 2 to the other side, we get
g x sin (theta) /2

This is a bonus question, and if i can get this right I might be able to improve my grade. Please Help

 

[v0id19]

You have said that the clue is that the formula is derived from

v^2 x sin 2 theta/g

I assume this means v^2 \cdot \frac{sin(2\theta)}{g}?

Also, a statement of the actual question would be very helpful.

 

[jhae2.718]

The range of a projectile is given by r = \frac{v_0^2\sin(2\theta)}{g}. (r \equiv s_h here.)

Could you post the exact wording of the question?

 

[arpita45]

present a formal derivation of the formula (given above) that was used to calculate the kinetic energy of the arrow immediately after release from the bow.. repeated here:

K = ½ x mgsH
This is the exact wording of the question. Hope it helps, also r = sh.