A bomb is flying horizontally at a heitht of ` 2000 m` with a speed of ` 200 ms^(-1)`. When a bomb is relesed from it . Find (i) the time taken by bomb to resch the ground(ii) the velocity with which the bomb hits the target and the distance of the target and the distance of the target from where the bomb is dropped. Take ` g=10 m//s^(-2)`.

Explanation

Let bomb be released from (O) and target be at (B) and bomb hit the target at (B) after time (t) fig. 2 (d) 39.
.
(i) Taking vertical downwrad motion of bomb from ` (O) to (B) we have
` u_y 0, y_0 =0, y= 2000 m , a-y = 10 ms^(-2)` , t= ?`
y=y_0 + u_y t + 1/2 a_y t^2`
:. 2000 =0 + 0 + 1/2 xx 10 t^2`
or ` t^2 = ( 400) /( 10) = 400 ` or ` t= 20 s` ltbRgt (ii) Horizontal velocityat (B),
` v_x =u_x =200 ms^(-1)` ,
Vertical velocity at (B),
` v-y = a_y t= 10 xx 20 = 200 ms^(-1)`
Velocity with which the bomb hits the target is
` v= sqrt (v_x^2 +v_y^2) = sqrt( (200)^2 + (200)^2)`
` = 200 sqrt 2 ms^(-1)`
` tan theta = v_y /v_x = ( 200) /( 200) = 1 ` or ` theta = 45^2`
` (iii) Horizontal distance ` AB = u_xt`
` = 200 xx 20 = 4000 m`
Distance of target rom ` O = OB`
` sqrt (( 200)^2 + 4000^2) = 2000 sqrt 5 m`.