A helicopter is in a level flight with an altitude of 4000 feet and a speed of 130 feet/sec. The helicopter drop a bomb with 0 speed relative to the helicopter. How far does the bomb travel horizontally from the launch point to impact with the ground?
how long does it take to drop 4000 ft?
use that time and 130 ft/s to get the distance needed.
1463.58 ft
To calculate the horizontal distance traveled by the bomb, we can use the concepts of time and velocity. We know that the horizontal distance traveled by the bomb is equal to the product of the time it takes to reach the ground and the horizontal velocity of the helicopter.
First, let's find the time it takes for the bomb to reach the ground. Since the helicopter is in level flight and the bomb is dropped with no initial velocity relative to the helicopter, we can use the formula for free fall distance:
h = (1/2) * g * t^2
where h is the altitude (4000 ft) and g is the acceleration due to gravity (approximately 32.2 ft/s^2).
Rearranging the formula to solve for t:
t^2 = (2h) / g
t^2 = (2 * 4000) / 32.2
t^2 = 495.34
t ≈ sqrt(495.34)
t ≈ 22.25 seconds
Now that we know the time it takes for the bomb to reach the ground, we can calculate the horizontal distance traveled using the horizontal velocity of the helicopter.
Distance = Velocity * Time
Distance = 130 ft/s * 22.25 s
Distance ≈ 2892.5 feet
Therefore, the bomb will travel approximately 2892.5 feet horizontally from the launch point to impact with the ground.