Answer: The distance is 583.1 m
Step-by-step 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