First Order ODE question on a projectile...

i need help with understanding this question - esp the part given in bold:

"In Dallas on November 22, 1963, President Kennedy was assassinated by Lee
Harvey Oswald (if you do not believe any of the conspiracy theories). Oswald
fired a Mannlicher-Carcano rifle from approximately 90 metres away. The
sight on Oswald’s rifle was less than ideal; if the bullet travelled in a straight
line after leaving the rifle (at a velocity of roughly 700 m/s) then the sight
aimed about 10 cm too high at a target 90 metres away. How much would the drop in the trajectory due to gravity compensate for this? (The initial vertical
velocity is zero, and satisfies dv/dt = -g, while the horizontal velocity is constant if we neglect air resistance.)"

I know its a projectile question using the ODE and equations of motion but the "sight" part is confusing me - help please!!

Reply 1

Zii

The sight is like the aimhole. Its where the gun was aiming.

Reply 2

zrancis

I did this question the other night, problem sheet 1?

It basically means the rifle shoots 10cm too high but you need to work out how much gravity will make it fall and if this is less than, equal to or more than the 10cm error. It's firing in a straight line.

Reply 3

ramdayal

i seem to get the drop due to gravity being 81g/9800 is this correct? - the thing is i dont really understand wat the question is askin for...

Reply 4

zrancis

Yep, take g=9.8 makes it nicer. Well, you'll find that the bullet without gravity would have hit 10cm higher, but with gravity it will actually be closer to the original target. I just put how much it falls by and the distance from the desired target as my answer.