A bomb is dropped from a plane flying at 4000 meters. The plane is flying at 80 m/s. How far, in a horizontal direction, did the bomb hit the ground?
S=(1/2)gt^2 since u=0
4000=(1/2)*10*t^2
t=sqrt(800)secs - time to reach ground
So hor. distance covered= 80*sqrt(800)
= 2263m
h = ho - o.5g*t^2 = 0
4000 - 4.9t^2 = 0
-4.9t^2 = -4000
t^2 = 816.3
Tf = 28.6 s. = Fall time or time in flight.
Dx = Hor = 80m/s * 28.6s = 2286 m.
To determine the horizontal distance traveled by the bomb, we'll need to use the formula for horizontal distance:
Distance = Speed × Time
First, we need to find the time it takes for the bomb to hit the ground. For a bomb dropped from a plane, we'll use the equations of motion for freefall:
Vertical distance = Initial velocity × Time + (1/2) × Acceleration × Time^2
The initial vertical velocity of the bomb is 0 because it is dropped, so the equation simplifies to:
Vertical distance = (1/2) × Acceleration × Time^2
Since we are only concerned with the time it takes for the bomb to hit the ground, we can set the vertical distance equal to the height at which the bomb was dropped (4000 meters):
4000 = (1/2) × 9.8 × Time^2
Simplifying the equation, we have:
9.8 × Time^2 = 8000
Now, solve for Time by dividing both sides of the equation by 9.8 and taking the square root:
Time = √(8000 / 9.8)
Time ≈ 12.65 seconds
Now that we know the time it takes for the bomb to hit the ground, we can calculate the horizontal distance using the formula:
Distance = Speed × Time
Distance = 80 m/s × 12.65 seconds
Distance ≈ 1012 meters
Therefore, the bomb hit the ground approximately 1012 meters horizontally from the point it was dropped.