When a aircraft is 4000m above the ground and flying upwards with a velocity of 50 m s-1 at an angle of 30° 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. (Assume g = 10 m s-2) The answer given by the book is: a) 31.0 s b) 264 m s-1, 9.5° from the vertical downwards How do you get those answers? Please help me. I have tried to get those answers but I only managed to get the first one. I used s=ut+1/2at^2 for the vertical component to get the time. Since u=0 for vertical component, -4000=-1/2(10)t^2 will get t=30.9s. Then for the second question, I got the horizontal velocity and vertical velocity when t=30.9s and using a vector diagram and the pythagoras theorem, I got -287 ms-1 for the velocity. However, the answer that was given by the book says it should be 264 m s-1. Is my answer correct or the book's answer correct?