Question

bomber 32000ft in air drops bomb. A projectile from canon 5000ft from point directly below bomber is fired five second after bomb is released. The projectile is set to intercept bomb at 1600ft.
(a) What is the initial speed of canon?
(b) What is the initial launch angle of projectile?

Answer 1

This is difficult :/

Answer 2

So the projectile must rise 1600 ft in five seconds less than the bomb drops 30400ft. We know how long it takes bombs to drop under gravity (s = u*t +1/2*a*t^2). So now we have time, t, until the collision. And we will be able to get a vertical component of the muzzle velocity, 1600/(t-5).

Alas, in order to get any sense out of the horizontal or the angle, we will need to know how fast the plane is flying, and in what direction - presumably towards the gun. Imagine a very slow aircraft v a very fast one. The bomb drops with the same initial forward speed as the bomber and this changes how far the shell has to go horizontally in the time we have available.

Also, I think that you may have an extra zero in there somewhere because in five seconds the bomb only falls about 125m = 400ft approx.