A bomb (JDAM) is released at t=0 with an initial velocity along the horizontal direction, vb = 200 m/s. A anti-aircraft system (LD-2000) fires immediately to intercept the bomb. The projectiles (from the gun) have an initial speed (muzzle velocity) of va = 1,000 m/s. The distance between the bomb and the anti-aircraft gun is shown in the graph. Neglect aerodynamic effects.(1) Determine the angle θ for the projectiles that allows successful interception.(2) Identify the time t’ and the height h at which the projectiles hit the bomb.

Sonu Saini
27 Points
one year ago
so,
let after t time bodies will collide.
relative velocity in horizontal direction is = (1000cosθ +200)m/s
distance will cover by this rel velocity is 5000m
so (1000cosθ +200)×t = 5000 ….eq 1
now reltive velolcity in y direction is =(1000sinθ)m/s
dis cover by this velolcity is=5000m
so (1000sinθ)×t = 5000  …..eq2
by eq1& eq2
1000cosθ + 200 = 1000sinθ
10cosθ +2 =10√{1-cos^2(θ)}
squaring
200cos^2(θ)  + 40cosθ = 100 – 4
25cos^2(θ)+5cosθ -12 = 0
after solving this quadratic eq we have two values of cosθ = 3/5 &-4/5 but -4/5 is not possible
so cosθ = 3/5
θ =53º