a plane flying horizontally @ 100 m/s @ 180 m above ground drops a bomb. How long will it be in air?

The plane is flying horizontally. Therefore the vertical component of its initial velocity is 0 m/s. It will therefore take the same amount of time to land as if it were just dropped 180 m. This time can be found using one of the kinematic equations of motion in the vertical direction, taking downwards to be positive. We have the following information:

u = initial velocity = 0 m/s

a = acceleration (due to gravity) = 9.8 m/s²

d = distance = 180 m

t = time = ?

d = ut + 1/2 at². Since u = 0, this reduces to d = 1/2 at² and rearranges to t = √(2d/a) = √(2*180/9.8) = 6.06 s, so it's in the air for 6.06 s.

Now how far away from the target must it be released. This is equal to the horizontal distance it travels in the 6.06 s it's in the air, and as the only force acting on it as it falls is gravity, which acts straight downwards, there are no horizontal forces acting on it, meaning no horizontal acceleration and therefore the horizontal component of velocity is constant, which means horizontal distance is horizontal speed times horizontal time

= (100 m/s)(6.06 s)

= 606 m.

a...the place's the preliminary drop .....don't understand b......c... i think from a undeniable top all issues attain terminal velocity without some type of thrust...d could that not be counted on the burden factor of the front and back of the gadget