Answers

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

This Is a Certified Answer

×
Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.
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
0
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.
0