.12 km.
To calculate the horizontal distance from the target that the supplies should be dropped, we can use the formula for projectile motion:
\(s = ut + 0.5at^2\),
where:
- s is the distance traveled horizontally,
- u is the initial horizontal velocity of the supplies (which is the same as the horizontal component of the airplane's velocity),
- a is the horizontal acceleration of the supplies (which is 0 since there is no horizontal force acting on the supplies once they are dropped),
- t is the time it takes for the supplies to hit the ground.
We can calculate the time it takes for the supplies to hit the ground using the formula:
\(s = ut + 0.5gt^2\),
where:
- s is the vertical distance the supplies are dropped (80 m in this case),
- u is the initial vertical velocity of the supplies (which is 0 since they are dropped),
- g is the acceleration due to gravity (9.81 m/s^2),
- t is the time it takes for the supplies to hit the ground.
Substitute the values into the equation:
\(80 = 0.5 * 9.81 * t^2\),
\(t = \sqrt{\frac{160}{9.81}} = 4.02\) (approximately).
Now, we can calculate the horizontal distance:
\(s = 350 \frac{km}{h} * \frac{1000m}{1km} * \frac{1h}{3600s} * 4.02 = 393.12m = 0.39312km\).
Therefore, the horizontal distance from the target that the supplies should be dropped is 0.39312 km or approximately 393 meters.
An airplane is flying level at 80 m above the ground with a speed of 350 km/h max wishes to drop food and medical supplies to hit a target on the ground I want horizontal distance from the target should max release the supplies The answer is 393
1 answer