An airplane flying horizontally 1000 m above the ground at 200km/h drops a cargo at a target on the ground.Determine where the cargo will strike the ground.
To determine where the cargo will strike the ground, we can break down the problem into two parts: one horizontal motion and one vertical motion.
First, let's consider the horizontal motion. The airplane is flying horizontally at a constant speed of 200 km/h. Since the cargo is dropped from the airplane, it will inherit the horizontal velocity of the airplane. Therefore, the horizontal velocity of the cargo is also 200 km/h.
Next, let's consider the vertical motion. The cargo is dropped from the airplane, so it will only be influenced by the force of gravity in the vertical direction. We can use the kinematic equation to determine the vertical distance traveled by the cargo.
The equation that relates distance, initial velocity, time, and acceleration (in this case, the acceleration due to gravity) is:
d = (1/2) * g * t^2
where:
d = vertical distance traveled
g = acceleration due to gravity (approximately 9.8 m/s^2)
t = time
Since the cargo is dropped, its initial vertical velocity is 0. Using this information, we can calculate the time it takes for the cargo to hit the ground.
1000 m = (1/2) * 9.8 m/s^2 * t^2
Simplifying the equation, we have:
t^2 = 200 / 9.8
t ≈ sqrt(20.41)
t ≈ 4.52 seconds
Now that we have the time it takes for the cargo to fall, we can calculate the horizontal distance traveled using the horizontal velocity.
distance = velocity * time
distance = 200 km/h * (4.52 s / 3600 s) (convert km/h to m/s)
distance ≈ 2.52 km
Therefore, the cargo will strike the ground approximately 2.52 kilometers away from the target point.
To determine where the cargo will strike the ground, we need to first find the time it takes for the cargo to reach the ground.
The horizontal speed of the airplane does not affect the vertical motion, so we can ignore it for now.
We can use the equation for vertical motion, which is given by:
s = ut + (1/2)gt^2
Where:
s = distance (in this case, the height of the cargo from the ground) = 1000 m
u = initial velocity (in this case, 0 m/s as the cargo is dropped)
t = time taken
g = acceleration due to gravity = 9.8 m/s^2 (approximate value)
Plugging in these values, the equation becomes:
1000 = 0*t + (1/2)*9.8*t^2
Simplifying the equation, we get:
4.9t^2 = 1000
Dividing both sides by 4.9, we get:
t^2 = 1000/4.9
Taking the square root of both sides, we get:
t ≈ √(1000/4.9)
Calculating this, we find:
t ≈ 6.42 seconds (approximately)
Therefore, it will take approximately 6.42 seconds for the cargo to reach the ground.
Now, to determine where the cargo will strike the ground horizontally, we can use the equation:
distance = speed * time
Where:
distance = horizontal distance (what we need to find)
speed = horizontal speed of the airplane = 200 km/h
time = 6.42 seconds
Converting the speed from km/h to m/s, we get:
speed = 200 km/h ≈ (200 * 1000) / (60 * 60) m/s ≈ 55.56 m/s
Plugging in these values, the equation becomes:
distance = 55.56 m/s * 6.42 s
Calculating this, we find:
distance ≈ 356.4 meters
Therefore, the cargo will strike the ground approximately 356.4 meters horizontally from the point where it was dropped.
how long does it take to fall 1000m?
4.9t^2 = 1000
The horizontal distance traveled is speed * time