The pilot of an airplane traveling 250km/h wants to drop supplies to flood victims isolated on a patch of land 190m below.
how long to drop 190 m?
190 = (1/2) (9.81)t^2
t = 6.22 seconds
how far horizontal in 6.22 s ?
(250,000 m/3600 s)6.22 s = 432 meters
so drop 432 meters before reaching target zenith
To calculate the time it will take for the supplies to reach the flood victims, we need to consider the horizontal distance the airplane covers and the vertical distance the supplies need to drop.
First, let's calculate the time it takes for the supplies to fall to the ground using the vertical distance. We can use the equation of motion for free-falling objects:
s = ut + (1/2)gt^2
Where:
- s is the vertical distance (190m)
- u is the initial vertical velocity (which is 0 since the supplies are dropped)
- g is the acceleration due to gravity (approximately 9.8 m/s^2)
- t is the time we want to find
Rearrange the equation and plug in the values:
190 = (1/2)(9.8)t^2
Simplifying the equation:
t^2 = (2 * 190) / 9.8
t^2 = 38.775
t ≈ √38.775
t ≈ 6.23 seconds (rounded to two decimal places)
So, it would take approximately 6.23 seconds for the supplies to reach the flood victims vertically.
Next, we need to calculate the horizontal distance the airplane covers in that time. We can use the equation:
distance = speed × time
Plugging in the values:
distance = 250 km/h × (6.23 seconds / 3600 seconds/hour)
distance = 0.44 km (rounded to two decimal places)
Therefore, the supplies will be dropped approximately 0.44 km horizontally from the starting point.
As a result, the pilot needs to ensure that the supplies are dropped at a horizontal distance of 0.44 km from the flood victims' location in order to reach them.