Question

a rifle shoots a bullet with a muzzle velocity of 400m/sec at a small a target 400 meter away. the height above the target at which the bullet must be aimed to hit the target is.(take g = ms-2)

Answer 1

so let the height H = $S_{y}$
no we know 
u = 400 m/s
$u_{y}$ = usinθ = 400sin0 = 0m/s
$u_{x}$ = ucosθ = 400cos0 = 400m/s
s = ut + 1/2at²
so 
$S_{x} = u_{x}t + 1/2a_{x}t^{2}$
⇒400 = 400t (along x axis g = 0m/s²)
⇒ t = 1 sec
again
$S_{y} = u_{y}t + 1/2g_{y}t^{2}$
H = 0 + 1/2g(1)²
H = 5 mtr
so it must be at a height 5m

Answer 2

If the angle of shooting is α
time taken to reach the target
t=400/(400* cosα)
  =1/cosα
vertical fall in this time = 1/2 * g* t²
                                 = 1/2 * 9.8 * 1/cos²α
tanα= vertical fall/distance
       =(1/2 *9,8*1/cos²α)/400
sinα * cos α =9.8/800
sin2α =2*9.8/800
          =0.0245
2α= sin inverse of 0.0245
    =1.404
α=0.702
tanα =0.01225
height of aim = tanα * 400
                    =0.01225*400
                    =4.9 meters above the horizontal.