Find the velocity and direction of the bomb just before it strikes the ground. Also determine the distance travelled by the plane before the bomb just strikes the ground.

Given

$u_x=200m/s \\ u_y=0 m/s \\ S_y=-400m \\ a_x=0 \\ a_y= -9.81m/s$

Horizontal

$V_y^2=u_y^2+2a_y.s_y \\ v_x=u_x=200m/s $

vertical

$V_x^2=u_x^2+2a_x.s_x \\ = 0+2(-9.81)(-400) \\ V_y = 88.589 m/s$

Velocity of bomb,

$V=\sqrt{V_x^2+V_y^2 } = \sqrt{200^2+88.589^2 } \\ V= 218.74 m/s \\ θ=\tan^{-1} (\dfrac {V_y}{V_x} )= \tan^{-1} (\dfrac {88.589}{200}) \\ θ=23.89^0\\ S_y=u_y t+1/2 a_y t^2 \\ -400 = 0 – \dfrac 12 \times 9.81 \times t^2 \\ t = 9.03 sec$

Distance travelled by plane in 9.03 sec

$s=ut+\dfrac 12 at^2 $

$ = 200 \times 9.03 + 0 $ … Acceleration in X.dir, $a_x=0$

$ S = 1806.09 m$