Object dropped from plane - where does it land?

[x^2]

Homework Statement

A plane is dropping an object from a height of 227m, such that it lands at a group of people on the ground. The speed of the plane is 60 m/s. How many meters before the group does the plane have to drop the object such that it lands at their position?

Homework Equations

x_f = x_i + v_ix delta t + (1/2) a_x(delta t)²

The Attempt at a Solution

From my understanding, I need to break up the x and y components of this problem. If I can find the time it takes the object to hit the ground, I can multiply it by the plane's velocity and see what distance the plane must be from the observers for the package to hit its target.

227 m = 9.8t m/s^2
(227 m)/(9.8 m/s^2) = t
t^2 = 23.162
t = sqrt(23.162)
t = 4.813 seconds

Now that I have the time it took for the object to fall, I can multiply it by the plane's velocity to see how far in the x direction the object will travel in 4.813 seconds:

4.813 s * 60 m/s = 288.78 m

This is the answer I get, but it isn't correct. I can't see how multiplying the velocity by the time is wrong, but perhaps I have the wrong time?

Thanks,
x^2

 

[Redbelly98]

227 m = 9.8t m/s^2

Something is wrong with that equation, since the units are different on the 2 sides:

m ≠ s ⋅ m/s^2
m ≠ m/s

 

[baba89]

Dear [Name],

Thank you for your question. I can provide a response to the problem you have presented.

First, let us break down the problem into its components. We have a plane dropping an object from a height of 227m, with a speed of 60 m/s. The object is intended to land at a group of people on the ground. We need to determine the distance from the group that the object needs to be dropped in order to hit its target.

To solve this problem, we can use the equation xf = xi + vix delta t + (1/2) ax(delta t)2, where xf is the final position, xi is the initial position, vix is the initial velocity in the x direction, delta t is the time, and ax is the acceleration in the x direction.

First, we need to determine the time it takes for the object to hit the ground. We can do this by using the equation xf = xi + vix delta t + (1/2) ax(delta t)2 and setting xf to 0 (since the object will hit the ground at 0m):

0 = 227m + 60 m/s * delta t + (1/2) * (-9.8 m/s^2) * (delta t)^2

Simplifying, we get:
0 = 227m + 60 m/s * delta t - 4.9 (delta t)^2

We can solve this quadratic equation using the quadratic formula:
(delta t)^2 - 12.24 (delta t) - 227 = 0
Solving for delta t, we get:
delta t = 11.86 seconds or delta t = -19.12 seconds

Since we are dealing with time, we can disregard the negative solution and use the positive solution of 11.86 seconds.

Now that we have the time it takes for the object to hit the ground, we can determine the distance it will travel in the x direction. Using the equation xf = xi + vix delta t, we get:

xf = 0 + 60 m/s * 11.86 seconds
xf = 711.6 m

Therefore, the plane needs to drop the object 711.6 meters away from the group in order for it to hit its target.

I hope this explanation helps. If you have any further questions or need clarification, please do not hesitate