Abomber flies horizontally with a speed of 367 m/s relative to the ground. the altitude of the bomber is 4020 m and the terrain is level. neglect the effects of air resistance. the acceleration of gravity is 9.8 m / s² . how far from the point vertically under the point of release does a bomb hit the ground? answer in units of m.

10512m

Explanation:

Using formula for Uniform Velocity v = D/t ;

where v = Velocity (367 m/s); D = Vertical distance (?); t = length of time of motion (?).

To find t we use the Kinematic equation D= v₀t + (1/2(at²));

where D = Distance traveled (4020m); v₀ = initial velocity (0); a = acceleration;

4020 = (0)t + (1/2(9.8)t²)

4020 = 1/2(9.8)t²

t² = 4020/(1/2(9.8))

t = √(4020/4.9).

back to our formula for Uniform Velocity v = D/t;

D =  vt

D = (367)(√(4020/4.9))

D ≈ 10512m

find the sum of the two horizontal vectors and the sum of the two vertical vectors

isoceles right angled triangle