Un avión que vuela a 70 m/s deja caer una caja de provisiones. ¿Qué distancia horizontal recorrerá la caja antes de tocar el suelo, 340 m más abajo?

The distance is 583.1 m

Explanation:

I will solve it in English, you can translate this if you need.

This question can be translated to:

A plane that flies at 70m/s drops a box of provisions.

What is the horizontal distance that the box travels before touching the ground, 340m under it?

I will solve it in English

The acceleration of the box is a(t) = -9.8m/s^2 j, the minus sign is because the gravity acceleration points downwards.

where j stands for the y-axis, and I for the x-axis.

the velocity of the box can be obtained by integrating that:

v(t) = (-9.8m/s^2)*tj + c*j + k*i

where c is the initial velocity in y, in this case is 0 because the box is dropped, and k is the initial velocity in x, in this case is 70m/s.

So v(t) = (-9.8m/s^2)*t*j +70m/s*i

integrating again, we can find the positions of x and y  

x(t) = 70m/s*t + C

y(t) = (-4.9m/s2)*t^2 + H

Where C and H are the initial positions in X and Y. Suppose that the initial position in X is 0, and the initial position in Y is 340m, so we have:

y(t) =  (-4.9m/s2)*t^2 + 340m

the box reaches the floor at the time where y(t) = 0 =  (-4.9m/s2)*t^2 + 340m

340m = (-4.9m/s2)*t^2

t = √(340/4.9) = 8.33s

Now, we put this time in the equation of X(t) to find the horizontal distance:

X(8.33s) = (70m/s)*8.33s = 583.1m

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .