I need Help!!! Physics Professors help me!?

I already answered this:

( x=81941.5m is not correct)

Assuming no air resistance?

time to fall 2400 m is

t = √(2h/g) = 22.1 seconds (a)

3704.4 km/h = 1029 m/s

1029 m/s x 22.1 s = 22700 meters (b)

vertical speed is

Vy = √(2gh) = 217 m/s

Vx = 1029 m/s

V = √(Vx² + Vy²)

An F-18 flying at 3704.4 km/h in a horizontal flight drops a bomb. if the aircraft is at an altitude of 2400.0m, how much time till the bomb hits the ground? how far in the horizontal direction does the bomb travel? how fast is the bomb moving right before hitting the ground?

As the bomb moves down, its velocity increases 9.8 m/s each second.

Distance = ½ * g * t^2

2400 = ½ * 9.8 * t^2

Time = (2400 ÷ 4.9)^0.5 = 22.1 s

During this time, the bomb moves horizontally at a constant speed

3704.4 km/h * 1000 m/km * 1 hr/3600s = 1029 m/s

Horizontal distance = 1029 * 22.1 = 22740.9 m

The vertical velocity = 9.8 * t = 9.8 * 22.1 = 216.58 m/s

The final velocity of the bomb = (Horizontal velocity^2 + Vertical velocity^2)^0.5

The final velocity of the bomb = (22.1^2 + 216.58^2)^0.5

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