# 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)

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by 133

2014-08-15T15:16:18+05:30

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So let the height H =
no we know
u = 400 m/s
= usinθ = 400sin0 = 0m/s
= ucosθ = 400cos0 = 400m/s
s = ut + 1/2at²
so

⇒400 = 400t (along x axis g = 0m/s²)
⇒ t = 1 sec
again

H = 0 + 1/2g(1)²
H = 5 mtr
so it must be at a height 5m
2014-08-15T17:49:21+05:30
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.