i have a question here about kinematics. 1. When a aircraft is 4000m above the ground and flying upwards with a velocity of 50 ms-¹ at an angle of 30º to the horizontal. A bomb is released.Neglecting air resistance, calculate (a)the time taken by the bomb to reach the ground, (b)the velocity of the bomb when it strikes the ground.(g=10 ms-²) (a)h=4000m vertical component of velocity=25.5 horizontal component of velocity=43.3 s=ut+1/2at² 4000=0+5t² t=28.5 v=u+at 25=0+10t t=2.5 t=28.5+2.5 t=31 is it correct? (b)vertical component of velocity=25.5 v=u+at =0+10(28.5) =285 ms-¹ vertical component of velocity=285 ms-¹-25.5 ms-¹=259.5 ms-¹ velocity of bomb, v=(259.5²+43.3²)½ =263 ms-¹ plz someone check the answer 4 me.TQ!