# Prove that Y = H -  gX^2/2v^2 then rearrange to v= X   sqRt g/2(H-Y)A bullet from a gun is fired horizontally with a velocity V, H metres above ground level. At a horizontal displacement of X m...

Prove that Y = H -  gX^2/2v^2

then rearrange to v= X   sqRt g/2(H-Y)

A bullet from a gun is fired horizontally with a velocity V, H metres above ground level. At a horizontal displacement of X m the bullet is Y m above ground level.

Would apprecate any help just spacedout

The bullet will fall with the acceleration of gravity.  What you must do is find out how far it falls in the time it takes to get to a horizontal displacement of X.

distance=velocity * time for the horizontal motion (ignoring air resistance)

So time = horizontal distance/velocity = X/V

For the vertical distance travelled, the formula is:

vertical distance = .5gt^2 where g=32 ft/sec^2 and t=X/V (as we found above)

so vertical distance = .5*g*X^2/V^2

so Y = H - (g/2)*X^2/V^2

so H - Y = (g/2)*X^2/V^2

so V^2 = g*X^2/(2*(H-Y))

so V = X *(sqrt (g/2(H-Y)))