Another Physics Question ASAP!?

Question

A shiny new sports car sits in the parking lot of a car dealership. Above is a cargo plane, flying horizontally at 50m/s. At the exact same moment the plane is 125m directly above the car, a heavy crate acedientally falls from its cargo doors. Relative to the car, where will the crate hit?

The answer is 250m away from the car.
How do you do it?

Answer 1

Assuming there is no drag, the problem is solved this way:

You care about the distance in the x-direction (horizontal), but the height of the plane (y-direction), and its speed govern that. We must solve the problem in 2 seperate directions to get your answer.

Y Direction:
How long does it take for the crate to hit the ground? Using the position equation, solving for t, we can find that out.

y(final) = y(initial) + v(initial) * t + 0.5 * a * t^2

y(final) is the final hieight of the crate when it hits the ground... 0m.
y(initial) is the height it started at... 125m
v(initial) is the speed it started with, IN THE DIRECTION OF THE DISTANCE TRAVELED. In this case, we are talking about verticle distance traveled. The key word in your question is "dropped". It was not thrown down, so the initial speed int he y-direction is 0m/s.
t is how long the crate fell (this is what we are solving for).
a is the accelerating in the direction of the distance traveled. So we use Earth's gravitational constant of -9.8m/s^2.
Dropping these values into the equation, we get:

0m = 125m + 0m/s * t + 0.5 * (-9.8m/s^2) * t ^2
Solving for t, we get 5s.


X Direction:
Now that we know how long it took to hit the ground, we can calculate how far in the x-direction it traveled during that 5s, using the same equation. This time, we solve for x(final)

x(final) = x(initial) + v(initial) * t + 0.5 * a * t^2

x(final) is the final location of the crate when it hits the ground... this is what we are solving for.
x(initial) is the location it started at. The key to this number lies in the wording of the question. The crate fell just as the plane past over the top of the car. We can call this location m
v(initial) is the speed it started with, IN THE DIRECTION OF THE DISTANCE TRAVELED. In this case, we are talking about horizontal distance traveled. The plane is traveling horizontally at 50m/s when the crates falls out, so it is also traveling at 50m/s, horizontally.
t is how long the crate fell (we just solved for this.. 5s)
a is the accelerating in the direction of the distance traveled. Gravity does not push anything to the side, so there is no acceleration in the horizontal direction. This number is 0m/s^2.

Dropping these values into the equation, we get:

x(final) = 0m + 50m/s * 5s + 0.5 * (0m/s^2) * 5s ^2
Solving for x(final), we get 250m.

Answer 2

The distance an accelerating object travels in time T is:

s = (1/2)aT^2

Since the crate is accelerating downwards by gravity, a = acceleration of gravity = 9.8 m/s^2

The plane is at 125m so s =125 and you can find T

125 = 4.9T^2 and T = 5.05 seconds

Since the crate will be moving at the same speed as the plane then it will travel at 50 m/s for 5.05 seconds.

Distance = 50*5.05 = 252.5 meters

Note: if you want a less precise answer people generally use 10 m/s^2 for gravity which would give a time of 5 seconds exactly and a distance of 250 meters

Answer 3

The side-to-side motion and vertical motion are independent.

First, compute how long it will take for the crate to hit the ground

d = 1/2 at^2, d = 125, a=10 (we're simplifying here)

125 = 1/2(10)t^2 =

125 = 5t^2
25 = t^2
5 = t

So it took five seconds for the crate to fall.

Now, the crate has velocity from the plane flying by. This is 50m/sec. It continues to have this as it falls.

5 secs * 50m/sec = 250 meters displacement.

Answer 4

1) how log fot the crate to hid the ground?

h= 1/2 a t^2
t^2 = 2h/a
t^2 = 2 * 125 / 9.8 = 25.51
t = 5.05 seconds

2) horizontal velocity = 50 m/s : for 5.05 seconds
x = v*t = 50 * 5.05 = 252 meters