The pilot of an airplane traveling 180km/h wants to drop supplies to flood victims isolated on a patch of land 160m below. The supplies should be dropped how many seconds before the plane is directly overhead?
h = Vo*t + 0.5g*t^2 = 160 m.
0 + 4.9t^2 = 160
t^2 = 32.65
t = 5.7 s. = Time to strike ground.
A dart is being thrown at a board that is 2.00 m away. When it hits the board it is 0.306 m below the point where it was aimed and it was thrown at 8.00 m/s. How long was the dart in the air?
Well, dropping supplies from a moving airplane is certainly a "high-flying" challenge! Let's do some calculations to keep things grounded.
First, we need to determine the time it takes for the supplies to fall 160 meters. We can use a classic physics formula for this: d = (1/2) * g * t^2, where "d" is the distance (160 meters), "g" is the acceleration due to gravity (9.8 m/s^2), and "t" is the time.
So, rearranging the formula, we get: t = sqrt((2 * d) / g).
Plugging in the values, we have: t = sqrt((2 * 160) / 9.8) = 4 seconds.
Now let's calculate how far the plane will travel in 4 seconds. Since the plane is traveling at a constant speed of 180 km/h, we need to convert it to meters per second: 180 km/h = (180,000 meters) / (3,600 seconds) = 50 meters/second.
Therefore, in 4 seconds, the plane will travel 4 * 50 = 200 meters.
Hence, the supplies should be dropped 200 meters before the plane is directly overhead the flood victims' patch of land.
Just make sure to have a sticker on the supplies that says, "Please catch carefully - contents may have been delivered by a clown pilot!"
To determine how many seconds before the plane is directly overhead the supplies should be dropped, we need to calculate the time it takes for the supplies to fall from the plane to the ground.
Gravity will accelerate the supplies downward at a rate of 9.8 m/s^2 (acceleration due to gravity).
First, let's convert the speed of the plane from km/h to m/s:
180 km/h * (1000 m / 1 km) * (1 h / 3600 s) = 50 m/s
Now, we can use the kinematic equation to calculate the time it takes for the supplies to fall:
h = 1/2 * g * t^2
Where:
h is the vertical distance (160 m)
g is the acceleration due to gravity (9.8 m/s^2)
t is the time
Rearranging the equation to solve for t:
t^2 = (2 * h) / g
Plugging in the given values:
t^2 = (2 * 160 m) / 9.8 m/s^2
t^2 = 32.65 s^2
Taking the square root of both sides, we get:
t ≈ 5.71 s
Therefore, the supplies should be dropped approximately 5.71 seconds before the plane is directly overhead.