A rescue plane carries a package of emergency supplies to be dropped at a point that is very close to the target. The plane travels with a velocity of 75m/s and flies 135m above the target. How far away (horizontally) from the target must the pilot release the package?
how long does it take to fall 135m? 4.9t^2 = 135, so 5.25 seconds
So, how far does the package move horizontally in that time?
distance = speed * time
Answer
To determine the horizontal distance the pilot must release the package, we can use the equations of motion.
Let's consider the time it takes for the package to reach the ground. The vertical distance traveled by the package is 135m. We can use the equation of motion for vertical displacement:
Δy = V_iy * t + (1/2) * a * t^2
Where:
Δy = vertical displacement (135m)
V_iy = initial vertical velocity (0 m/s since the package is dropped)
a = acceleration due to gravity (-9.8 m/s^2)
t = time taken
Plugging in the values, we have:
135m = 0 * t + (1/2) * (-9.8 m/s^2) * t^2
270m = 9.8 m/s^2 * t^2
t^2 = 270m / (9.8 m/s^2)
t^2 = 27.551
Taking the square root of both sides, we find:
t ≈ 5.251 seconds (rounded to three decimal places)
Now, let's calculate the horizontal distance traveled by the package during this time. The plane's velocity is 75 m/s, so the horizontal distance (d) is given by:
d = V_ix * t
Where:
V_ix = initial horizontal velocity (75 m/s)
t = time taken (5.251 seconds)
Plugging in the values, we have:
d = 75 m/s * 5.251 s
d ≈ 393.825 meters (rounded to three decimal places)
Therefore, the pilot must release the package approximately 393.825 meters horizontally away from the target.
To find the horizontal distance from the target where the pilot must release the package, we can use the information given.
Let's break down the information we have:
- The velocity of the plane: 75 m/s
- The altitude of the plane above the target: 135 m
We can use the concept of time to find the horizontal distance. The time it takes for the package to fall vertically from the plane to the target is the same as the time it takes for the plane to travel horizontally to that point.
First, let's calculate the time it takes for the package to fall from the plane to the target. We can use the equation of motion:
s = u*t + (1/2)*a*t^2
In this case, u (initial velocity in the vertical direction) is 0 because the package is only falling due to gravity, and a (acceleration in the vertical direction) is the acceleration due to gravity, which is approximately 9.8 m/s^2.
Since the initial vertical velocity is 0, the equation simplifies to:
s = (1/2)*a*t^2
where s is the vertical distance (135 m) and t is the time taken.
135 = (1/2)*9.8*t^2
Solving for t:
t^2 = (135 * 2) / 9.8
t^2 = 27.55
t ≈ √27.55
t ≈ 5.25 seconds
Now that we have the time, we can find the horizontal distance traveled by the plane during this time.
Distance = velocity * time
Distance = 75 m/s * 5.25 s
Distance ≈ 393.75 meters
Therefore, the pilot must release the package approximately 393.75 meters horizontally away from the target.