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

the answer would be 32*f

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