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

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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)))

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